Solutions control systems engineering by norman nice 6 ed

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  • 1. Apago PDF EnhancerSOLUTIONMANUAL
  • 2. 1-17 Solutions to ProblemsCopyright © 2011 by John Wiley & Sons, Inc.Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understandingfor more than 200 years, helping people around the world meet their needs and fulfill theiraspirations. Our company is built on a foundation of principles that include responsibility to thecommunities we serve and where we live and work. In 2008, we launched a Corporate CitizenshipInitiative, a global effort to address the environmental, social, economic, and ethical challenges weface in our business. Among the issues we are addressing are carbon impact, paper specifications andprocurement, ethical conduct within our business and among our vendors, and community andcharitable support. For more information, please visit our website: www.wiley.com/go/citizenship.Copyright © 2011 by John Wiley & Sons, Inc.No part of this publication may be reproduced, stored in a retrieval system or transmitted in any formor by any means, electronic, mechanical, photocopying recording, scanning or otherwise, except aspermitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either theprior written permission of the Publisher or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Requests to the Publisher for permission should be addressed to thePermissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201) 748-6008 or online at http://www.wiley.com/go/permissions.Evaluation copies are provided to qualified academics and professionals for review purposes only, foruse in their courses during the next academic year. These copies are licensed and may not be sold ortransferred to a third party. Upon completion of the review period, please return the evaluation copyto Wiley. Return instructions and a free of charge return shipping label are available atwww.wiley.com/go/returnlabel. Outside of the United States, please contact your localrepresentativeISBN 13 978-0470-54756-4Student companion website
  • 3. Copyright © 2011 by John Wiley & Sons, Inc.O N EIntroductionANSWERS TO REVIEW QUESTIONS1. Guided missiles, automatic gain control in radio receivers, satellite tracking antenna2. Yes - power gain, remote control, parameter conversion; No - Expense, complexity3. Motor, low pass filter, inertia supported between two bearings4. Closed-loop systems compensate for disturbances by measuring the response, comparing it tothe input response (the desired output), and then correcting the output response.5. Under the condition that the feedback element is other than unity6. Actuating signal7. Multiple subsystems can time share the controller. Any adjustments to the controller can beimplemented with simply software changes.8. Stability, transient response, and steady-state error9. Steady-state, transient10. It follows a growing transient response until the steady-state response is no longer visible. Thesystem will either destroy itself, reach an equilibrium state because of saturation in drivingamplifiers, or hit limit stops.11. Natural response12. Determine the transient response performance of the system.13. Determine system parameters to meet the transient response specifications for the system.14. True15. Transfer function, state-space, differential equations16. Transfer function - the Laplace transform of the differential equationState-space - representation of an nth order differential equation as n simultaneous first-orderdifferential equationsDifferential equation - Modeling a system with its differential equationSOLUTIONS TO PROBLEMS1. Five turns yields 50 v. Therefore K =50 volts5 x 2π rad= 1.59
  • 4. 1-2 Chapter 1: IntroductionCopyright © 2011 by John Wiley & Sons, Inc.2.ThermostatAmplifier andvalvesHeaterTemperaturedifferenceVoltagedifferenceFuelflowActualtemperatureDesiredtemperature+-3.DesiredrollangleInputvoltage+-PilotcontrolsAileronpositioncontrolErrorvoltageAileronpositionAircraftdynamicsRollrateIntegrateRollangleGyroGyro voltage4.SpeedErrorvoltageDesiredspeedInputvoltage+-transducer AmplifierMotoranddrivesystemActualspeedVoltageproportionalto actual speedDancerpositionsensorDancerdynamics
  • 5. 1-3 Solutions to ProblemsCopyright © 2011 by John Wiley & Sons, Inc.5.DesiredpowerPowerErrorvoltageInputvoltage+-Transducer AmplifierMotoranddrivesystemVoltageproportionalto actual powerRodpositionReactorActualpowerSensor &transducer6.Desiredstudentpopulation +-AdministrationPopulationerrorDesiredstudentrateAdmissionsActualstudentrate +-Graduatinganddrop-outrateNet rateof influxIntegrateActualstudentpopulation7.Desiredvolume +-TransducerVolumecontrol circuitVoltageproportionalto desiredvolumeVolumeerrorRadioVoltagerepresentingactual volume Actualvolume-+Transducer-SpeedVoltageproportionalto speedEffectivevolume
  • 6. 1-4 Chapter 1: IntroductionCopyright © 2011 by John Wiley & Sons, Inc.8.a.R+V-VDifferentialamplifierDesiredlevel+-PoweramplifierActuatorValveFloatFluid inputDrainTankR+V-Vb.DesiredlevelAmplifiers Actuatorand valveFlowrate inIntegrateActuallevelFlowrate outPotentiometer+-+DrainFloatPotentiometer-voltageinvoltageoutDisplacement
  • 7. 1-5 Solutions to ProblemsCopyright © 2011 by John Wiley & Sons, Inc.9.DesiredforceTransducer Amplifier Valve Actuatorand loadTireLoad cellActualforce+-Current Displacement Displacement10.Commandedblood pressureVaporizer PatientActualbloodpressure+-Isofluraneconcentration11.+-Controller&motorGrinderForce Feed rateIntegratorDesireddepth Depth12.+-CoilcircuitSolenoid coil& actuatorCoilcurrent Force Armature&spool dynamicsDesiredposition DepthTransducerCoilvoltageLVDT
  • 8. 1-6 Chapter 1: IntroductionCopyright © 2011 by John Wiley & Sons, Inc.13.a.b.If the narrow light beam is modulated sinusoidally the pupil’s diameter will alsovary sinusoidally (with a delay see part c) in problem)c. If the pupil responded with no time delay the pupil would contract only to the pointwhere a small amount of light goes in. Then the pupil would stop contracting andwould remain with a fixed diameter.+DesiredLightIntensityBrain Internal eyemusclesRetina + OpticalRetina’sLightIntensityNervoussystemelectricalimpulsesNervoussystemelectricalimpulses+DesiredLightIntensityBrain Internal eyemusclesRetina + OpticalNervesRetina’sLightIntensityExternalLight
  • 9. 1-7 Solutions to ProblemsCopyright © 2011 by John Wiley & Sons, Inc.14.15.16.17.a. Ldidt+ Ri = u(t)b. Assume a steady-state solution iss = B. Substituting this into the differential equation yields RB =1,from which B =1R. The characteristic equation is LM + R = 0, from which M = -RL. Thus, the total+Desired AmplifierGyroscopicActualHT’s
  • 10. 1-8 Chapter 1: IntroductionCopyright © 2011 by John Wiley & Sons, Inc.solution is i(t) = Ae-(R/L)t +1R. Solving for the arbitrary constants, i(0) = A +1R= 0. Thus, A =-1R. The final solution is i(t) =1R--1Re-(R/L)t =1R(1 − e−( R/ L)t).c.18.a. Writing the loop equation, Ri + Ldidt+1Cidt + vC (0)∫ = v(t)b. Differentiating and substituting values,222 25 0d i diidt dt+ + =Writing the characteristic equation and factoring,22 25 ( 1 24 )( 1 24 )M M M i M i+ + = + + + − .The general form of the solution and its derivative iscos( 24 ) sin( 24 )t ti Ae t Be t− −= +( 24 ) cos( 24 ) ( 24 ) sin( 24 )t tdiA B e t A B e tdt− −= − + − +Using(0) 1(0) 0; (0) 1Ldi vidt L L= = = =i 0 A= =0(0) 24diA Bdt= − + =1Thus, A = 0 and124B = .The solution is1sin( 24 )24ti e t−=
  • 11. 1-9 Solutions to ProblemsCopyright © 2011 by John Wiley & Sons, Inc.c.19.a. Assume a particular solution ofSubstitute into the differential equation and obtainEquating like coefficients,From which, C =3553 and D =1053 .The characteristic polynomial isThus, the total solution isSolving for the arbitrary constants, x(0) = A +3553 = 0. Therefore, A = -3553 . The final solution isb. Assume a particular solution of
  • 12. 1-10 Chapter 1: IntroductionCopyright © 2011 by John Wiley & Sons, Inc.xp = Asin3t + Bcos3tSubstitute into the differential equation and obtain(18A − B)cos(3t) − (A +18B)sin(3t) = 5sin(3t)Therefore, 18A – B = 0 and –(A + 18B) = 5. Solving for A and B we obtainxp = (-1/65)sin3t + (-18/65)cos3tThe characteristic polynomial isM2+ 6 M + 8 = M + 4 M + 2Thus, the total solution isx = C e- 4 t+ D e- 2 t+ -1865cos 3 t -165sin 3 tSolving for the arbitrary constants, x(0) = C + D −1865= 0 .Also, the derivative of the solution is= -365cos 3 t +5465sin 3 t - 4 C e- 4 t- 2 D e- 2 tdxdtSolving for the arbitrary constants, x.(0) −365− 4C − 2D = 0 , or C = −310and D =1526.The final solution isx = -1865cos 3 t -165sin 3 t -310e- 4 t+1526e- 2 tc. Assume a particular solution ofxp = ASubstitute into the differential equation and obtain 25A = 10, or A = 2/5.The characteristic polynomial isM2+ 8 M + 25 = M + 4 + 3 i M + 4 - 3 iThus, the total solution isx =25+ e- 4 tB sin 3 t + C cos 3 tSolving for the arbitrary constants, x(0) = C + 2/5 = 0. Therefore, C = -2/5. Also, the derivative of thesolution is
  • 13. 1-11 Solutions to ProblemsCopyright © 2011 by John Wiley & Sons, Inc.= 3 B -4 C cos 3 t - 4 B + 3 C sin 3 t e- 4 tdxdtSolving for the arbitrary constants, x.(0) = 3B – 4C = 0. Therefore, B = -8/15. The final solution isx(t) =25− e−4t 815sin(3t) +25cos(3t)⎛⎝⎞⎠20.a. Assume a particular solution ofSubstitute into the differential equation and obtainEquating like coefficients,From which, C = -15 and D = -110 .The characteristic polynomial isThus, the total solution isSolving for the arbitrary constants, x(0) = A -15 = 2. Therefore, A =115. Also, the derivative of thesolution isdxdtSolving for the arbitrary constants, x.(0) = - A + B - 0.2 = -3. Therefore, B = −35. The final solutionisx(t) = −15cos(2t) −110sin(2t) + e−t 115cos(t) −35sin(t)⎛⎝⎞⎠b. Assume a particular solution ofxp = Ce-2t + Dt + ESubstitute into the differential equation and obtain
  • 14. 1-12 Chapter 1: IntroductionCopyright © 2011 by John Wiley & Sons, Inc.Equating like coefficients, C = 5, D = 1, and 2D + E = 0.From which, C = 5, D = 1, and E = - 2.The characteristic polynomial isThus, the total solution isSolving for the arbitrary constants, x(0) = A + 5 - 2 = 2 Therefore, A = -1. Also, the derivative of thesolution isdxdt= (−A+ B)e− t− Bte−t−10e−2t+1Solving for the arbitrary constants, x.(0) = B - 8 = 1. Therefore, B = 9. The final solution isc. Assume a particular solution ofxp = Ct2 + Dt + ESubstitute into the differential equation and obtainEquating like coefficients, C =14 , D = 0, and 2C + 4E = 0.From which, C =14 , D = 0, and E = -18 .The characteristic polynomial isThus, the total solution isSolving for the arbitrary constants, x(0) = A -18 = 1 Therefore, A =98 . Also, the derivative of thesolution isdxdtSolving for the arbitrary constants, x.(0) = 2B = 2. Therefore, B = 1. The final solution is
  • 15. 1-13 Solutions to ProblemsCopyright © 2011 by John Wiley & Sons, Inc.21.+-InputtransducerDesiredforceInputvoltageController Actuator Pantographdynamics SpringFupSpringdisplacementFoutSensor22.Amount ofHIV virusesRTIPIDesiredAmount ofHIV virusesController Patient
  • 16. 1-14 Chapter 1: IntroductionCopyright © 2011 by John Wiley & Sons, Inc.23.a.SpeedActualMotiveForceECUVehicleElectricMotorAerodynamicClimbing &RollingResistancesAerodynamicSpeed+ +InverterControlCommandControlledVoltageInverterDesired
  • 17. 1-15 Solutions to ProblemsCopyright © 2011 by John Wiley & Sons, Inc.b.Desired Speed ActualMotiveECUAcceleratorDisplacementVehicleAccelerator,AerodynamicClimbing &RollingResistancesAerodynamicSpeed++_
  • 18. 1-16 Chapter 1: IntroductionCopyright © 2011 by John Wiley & Sons, Inc.c.SpeedErrorActualTotalMotiveForceECU VehicleAerodynamicClimbing &RollingResistancesAerodynamicSpeed++PowerPlanetaryGearControlInverterControlCommandInverter&ElectricMotorMotorAcceleratorAcceleratorICEDesired
  • 19. Copyright © 2011 by John Wiley & Sons, Inc.T W OModeling in theFrequency DomainSOLUTIONS TO CASE STUDIES CHALLENGESAntenna Control: Transfer FunctionsFinding each transfer function:Pot:Vi(s)θi(s)=10π;Pre-Amp:Vp(s)Vi(s) = K;Power Amp:Ea(s)Vp(s) =150s+150Motor: Jm = 0.05 + 5(50250 )2= 0.25Dm =0.01 + 3(50250 )2= 0.13KtRa=15KtKbRa=15Therefore:θm(s)Ea(s) =KtRaJms(s+1Jm(Dm+KtKbRa))=0.8s(s+1.32)And:θo(s)Ea(s) =15θm(s)Ea(s) =0.16s(s+1.32)Transfer Function of a Nonlinear Electrical NetworkWriting the differential equation,d(i0 + δi)dt+ 2(i0 +δi)2− 5 = v(t) . Linearizing i2about i0,(i0+δi)2- i02= 2i ⎮i=i0δi = 2i0δi.. Thus, (i0+δi)2= i02+ 2i0δi.
  • 20. Chapter 2: Modeling in the Frequency Domain 2-2Copyright © 2011 by John Wiley & Sons, Inc.Substituting into the differential equation yields,dδidt + 2i02 + 4i0δi - 5 = v(t). But, theresistor voltage equals the battery voltage at equilibrium when the supply voltage is zero sincethe voltage across the inductor is zero at dc. Hence, 2i02 = 5, or i0 = 1.58. Substituting into the linearizeddifferential equation,dδidt + 6.32δi = v(t). Converting to a transfer function,δi(s)V(s) =1s+6.32 . Using thelinearized i about i0, and the fact that vr(t) is 5 volts at equilibrium, the linearized vr(t) is vr(t) = 2i2 =2(i0+δi)2 = 2(i02+2i0δi) = 5+6.32δi. For excursions away from equilibrium, vr(t) - 5 = 6.32δi = δvr(t).Therefore, multiplying the transfer function by 6.32, yields,δVr(s)V(s) =6.32s+6.32 as the transfer functionabout v(t) = 0.ANSWERS TO REVIEW QUESTIONS1. Transfer function2. Linear time-invariant3. Laplace4. G(s) = C(s)/R(s), where c(t) is the output and r(t) is the input.5. Initial conditions are zero6. Equations of motion7. Free body diagram8. There are direct analogies between the electrical variables and components and the mechanical variablesand components.9. Mechanical advantage for rotating systems10. Armature inertia, armature damping, load inertia, load damping11. Multiply the transfer function by the gear ratio relating armature position to load position.12. (1) Recognize the nonlinear component, (2) Write the nonlinear differential equation, (3) Select theequilibrium solution, (4) Linearize the nonlinear differential equation, (5) Take the Laplace transform ofthe linearized differential equation, (6) Find the transfer function.SOLUTIONS TO PROBLEMS1.a. F(s) = e− stdt0∞∫ = −1se−st0∞=1sb. F(s) = te− stdt0∞∫ =e−sts2 (−st −1) 0∞=−(st +1)s2est0∞
  • 21. Solutions to Problems 2-3Copyright © 2011 by John Wiley & Sons, Inc.Using LHopitals RuleF(s)t→ ∞ =−ss3estt→∞= 0. Therefore, F(s) =1s2 .c. F(s) = sinωt e− stdt0∞∫ =e− sts2+ ω2 (−ssinωt − ωcosωt)0∞=ωs2+ω2d. F(s) = cosωt e− stdt0∞∫ =e− sts2+ ω2 (−scosωt + ωsinωt)0∞=ss2+ω22.a. Using the frequency shift theorem and the Laplace transform of sin ωt, F(s) =ω(s+a)2+ω2 .b. Using the frequency shift theorem and the Laplace transform of cos ωt, F(s) =(s+a)(s+a)2+ω2 .c. Using the integration theorem, and successively integrating u(t) three times, ⌡⌠dt = t; ⌡⌠tdt =t22 ;⌡⌠t22dt =t36 , the Laplace transform of t3u(t), F(s) =6s4 .3.a. The Laplace transform of the differential equation, assuming zero initial conditions,is, (s+7)X(s) =5ss2+22 . Solving for X(s) and expanding by partial fractions,Or,Taking the inverse Laplace transform, x(t) = -3553 e-7t + (3553 cos 2t +1053 sin 2t).b. The Laplace transform of the differential equation, assuming zero initial conditions, is,(s2+6s+8)X(s) =15s2+ 9.Solving for X(s)X(s) =15(s2+ 9)(s2+ 6s + 8)and expanding by partial fractions,X(s) = −3656s +199s2+ 9−3101s + 4+15261s + 2
  • 22. Chapter 2: Modeling in the Frequency Domain 2-4Copyright © 2011 by John Wiley & Sons, Inc.Taking the inverse Laplace transform,x(t) = −1865cos(3t) −165sin(3t) −310e−4t+1526e−2tc. The Laplace transform of the differential equation is, assuming zero initial conditions,(s2+8s+25)x(s) =10s. Solving for X(s)X s =10s s2+ 8 s + 25and expanding by partial fractions,X s =251s-251 s + 4 +499s + 42+ 9Taking the inverse Laplace transform,x(t) =25− e−4t 815sin(3t) +25cos(3t)⎛⎝⎞⎠4.a. Taking the Laplace transform with initial conditions, s2X(s)-4s+4+2sX(s)-8+2X(s) =2s2+22 .Solving for X(s),X(s) =3 22 24 4 16 18( 4)( 2 2)s s ss s s+ + ++ + +.Expanding by partial fractions2 2 21s 21 1 21(s 1) 22X(s)5 s 2 5 (s 1) 1++ +⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠Therefore,1 2 1( ) 21 cos sin sin 2 cos25 21 2t tx t e t e t t t− −⎡ ⎤= + − −⎢ ⎥⎣ ⎦b. Taking the Laplace transform with initial conditions, s2X(s)-4s-1+2sX(s)-8+X(s) =5s+2 +1s2 .Solving for X(s),4 3 22 24 17 23 2( )( 1) ( 2)s s s sX ss s s+ + + +=+ +2 21 2 11 1 5( )( 1) ( 1) ( 2)X ss s s s s= − + + ++ + +Therefore2( ) 2 11 5t t tx t t te e e− − −= − + + + .c. Taking the Laplace transform with initial conditions, s2X(s)-s-2+4X(s) =2s3 . Solving for X(s),
  • 23. Solutions to Problems 2-5Copyright © 2011 by John Wiley & Sons, Inc.4 33 22 3 2( )( 4)s sX ss s+ +=+2 317 3*21/ 2 1/88 2( )4sX ss s s+= + −+Therefore217 3 1 1( ) cos2 sin 28 2 4 8x t t t t= + + − .5.Program:syms tatheta=45*pi/180f=8*t^2*cos(3*t+theta);pretty(f)F=laplace(f);F=simple(F);pretty(F)btheta=60*pi/180f=3*t*exp(-2*t)*sin(4*t+theta);pretty(f)F=laplace(f);F=simple(F);pretty(F)Computer response:ans =atheta =0.78542 / PI 8 t cos| -- + 3 t | 4 /1/2 28 2 (s + 3) (s - 12 s + 9)------------------------------2 3(s + 9)ans =
  • 24. Chapter 2: Modeling in the Frequency Domain 2-6Copyright © 2011 by John Wiley & Sons, Inc.btheta =1.0472/ PI 3 t sin| -- + 4 t | exp(-2 t) 3 /1/2 21/2 1/2 3 3 s12 s + 6 3 s - 18 3 + --------- + 242------------------------------------------2 2(s + 4 s + 20)6.Program:syms saG=(s^2+3*s+10)*(s+5)/[(s+3)*(s+4)*(s^2+2*s+100)];pretty(G)g=ilaplace(G);pretty(g)bG=(s^3+4*s^2+2*s+6)/[(s+8)*(s^2+8*s+3)*(s^2+5*s+7)];pretty(G)g=ilaplace(G);pretty(g)Computer response:ans =a2(s + 5) (s + 3 s + 10)--------------------------------2(s + 3) (s + 4) (s + 2 s + 100)/ 1/2 1/2 | 1/2 11 sin(3 11 t) |5203 exp(-t) | cos(3 11 t) - -------------------- |20 exp(-3 t) 7 exp(-4 t) 57233 /------------ - ----------- + ------------------------------------------------------103 54 5562
  • 25. Solutions to Problems 2-7Copyright © 2011 by John Wiley & Sons, Inc.ans =b3 2s + 4 s + 2 s + 6-------------------------------------2 2(s + 8) (s + 8 s + 3) (s + 5 s + 7)/ 1/2 1/2 | 1/2 4262 13 sinh(13 t) |1199 exp(-4 t) | cosh(13 t) - ------------------------ | 15587 /----------------------------------------------------------- -417/ / 1/2 | 1/2 | 3 t | || / 1/2 131 3 sin| ------ | |/ 5 t | | 3 t | 2 / |65 exp| - --- | | cos| ------ | + ---------------------- | 2 / 2 / 15 / 266 exp(-8 t)---------------------------------------------------------- - -------------4309 937.The Laplace transform of the differential equation, assuming zero initial conditions, is,(s3+3s2+5s+1)Y(s) = (s3+4s2+6s+8)X(s).Solving for the transfer function,Y(s)X(s)=s3+ 4s2+ 6s + 8s3+3s2+ 5s +1.8.a. Cross multiplying, (s2+5s+10)X(s) = 7F(s).Taking the inverse Laplace transform,d2xdt2 + 5dxdt+ 10x = 7f.b. Cross multiplying after expanding the denominator, (s2+21s+110)X(s) = 15F(s).Taking the inverse Laplace transform,d2xdt2 + 21dxdt+ 110x =15f.c. Cross multiplying, (s3+11s2+12s+18)X(s) = (s+3)F(s).Taking the inverse Laplace transform,d3xdt3 + 11d2xdt2 + 12dxdt+ 18x = dft/dt + 3f.9.The transfer function isC(s)R(s)=5 4 3 26 5 4 3 22 4 47 3 2 5s s s ss s s s s+ + + ++ + + + +.Cross multiplying, (s6+7s5+3s4+2s3+s2+5)C(s) = (s5+2s4+4s3+s2+4)R(s).Taking the inverse Laplace transform assuming zero initial conditions,d6cdt6 + 7d5cdt5 + 3d 4cdt4 + 2d3cdt3 +d2cdt2 + 5c =d5rdt5 + 2d 4rdt4 + 4d3rdt3 +d2rdt2 + 4r.10.The transfer function isC(s)R(s)=s4+ 2s3+ 5s2+ s +1s5+ 3s4+ 2s3+ 4s2+ 5s + 2.
  • 26. Chapter 2: Modeling in the Frequency Domain 2-8Copyright © 2011 by John Wiley & Sons, Inc.Cross multiplying, (s5+3s4+2s3+4s2+5s+2)C(s) = (s4+2s3+5s2+s+1)R(s).Taking the inverse Laplace transform assuming zero initial conditions,d5cdc5 + 3d 4cdt4 + 2d3cdt3 + 4d2cdt2 + 5dcdt+ 2c =d 4rdt4 + 2d3rdt3 + 5d2rdt2 +drdt+ r.Substituting r(t) = t3,d5cdc5 + 3d 4cdt4 + 2d3cdt3 + 4d2cdt2 + 5dcdt+ 2c= 18δ(t) + (36 + 90t + 9t2+ 3t3) u(t).11.Taking the Laplace transform of the differential equation, s2X(s)-s+1+2sX(s)-2+3x(s)=R(s).Collecting terms, (s2+2s+3)X(s) = R(s)+s+1.Solving for X(s), X(s) =R(s)s2+ 2s +3+s +1s2+ 2s +3.The block diagram is shown below, where R(s) = 1/s.12.Program:FactoredGzpk=zpk([-15 -26 -72],[0 -55 roots([1 5 30]) roots([1 27 52])],5)PolynomialGp=tf(Gzpk)Computer response:ans =FactoredZero/pole/gain:5 (s+15) (s+26) (s+72)--------------------------------------------s (s+55) (s+24.91) (s+2.087) (s^2 + 5s + 30)ans =PolynomialTransfer function:5 s^3 + 565 s^2 + 16710 s + 140400--------------------------------------------------------------------s^6 + 87 s^5 + 1977 s^4 + 1.301e004 s^3 + 6.041e004 s^2 + 8.58e004 s13.Program:PolynomialGtf=tf([1 25 20 15 42],[1 13 9 37 35 50])
  • 27. Solutions to Problems 2-9Copyright © 2011 by John Wiley & Sons, Inc.FactoredGzpk=zpk(Gtf)Computer response:ans =PolynomialTransfer function:s^4 + 25 s^3 + 20 s^2 + 15 s + 42-----------------------------------------s^5 + 13 s^4 + 9 s^3 + 37 s^2 + 35 s + 50ans =FactoredZero/pole/gain:(s+24.2) (s+1.35) (s^2 - 0.5462s + 1.286)------------------------------------------------------(s+12.5) (s^2 + 1.463s + 1.493) (s^2 - 0.964s + 2.679)14.Program:numg=[-5 -70];deng=[0 -45 -55 (roots([1 7 110])) (roots([1 6 95]))];[numg,deng]=zp2tf(numg,deng,1e4);Gtf=tf(numg,deng)G=zpk(Gtf)[r,p,k]=residue(numg,deng)Computer response:Transfer function:10000 s^2 + 750000 s + 3.5e006-------------------------------------------------------------------------------s^7 + 113 s^6 + 4022 s^5 + 58200 s^4 + 754275 s^3 + 4.324e006 s^2 + 2.586e007 sZero/pole/gain:10000 (s+70) (s+5)------------------------------------------------s (s+55) (s+45) (s^2 + 6s + 95) (s^2 + 7s + 110)r =-0.00180.00660.9513 + 0.0896i0.9513 - 0.0896i-1.0213 - 0.1349i-1.0213 + 0.1349i0.1353p =-55.0000-45.0000-3.5000 + 9.8869i-3.5000 - 9.8869i-3.0000 + 9.2736i-3.0000 - 9.2736i0k =[]
  • 28. Chapter 2: Modeling in the Frequency Domain 2-10Copyright © 2011 by John Wiley & Sons, Inc.15.Program:syms s(a)Ga=45*[(s^2+37*s+74)*(s^3+28*s^2+32*s+16)].../[(s+39)*(s+47)*(s^2+2*s+100)*(s^3+27*s^2+18*s+15)];Ga symbolicpretty(Ga)[numga,denga]=numden(Ga);numga=sym2poly(numga);denga=sym2poly(denga);Ga polynimialGa=tf(numga,denga)Ga factoredGa=zpk(Ga)(b)Ga=56*[(s+14)*(s^3+49*s^2+62*s+53)].../[(s^2+88*s+33)*(s^2+56*s+77)*(s^3+81*s^2+76*s+65)];Ga symbolicpretty(Ga)[numga,denga]=numden(Ga);numga=sym2poly(numga);denga=sym2poly(denga);Ga polynimialGa=tf(numga,denga)Ga factoredGa=zpk(Ga)Computer response:ans =(a)ans =Ga symbolic2 3 2(s + 37 s + 74) (s + 28 s + 32 s + 16)45 -----------------------------------------------------------2 3 2(s + 39) (s + 47) (s + 2 s + 100) (s + 27 s + 18 s + 15)ans =Ga polynimialTransfer function:45 s^5 + 2925 s^4 + 51390 s^3 + 147240 s^2 + 133200 s + 53280--------------------------------------------------------------------------------s^7 + 115 s^6 + 4499 s^5 + 70700 s^4 + 553692 s^3 + 5.201e006 s^2 + 3.483e006 s+ 2.75e006ans =Ga factoredZero/pole/gain:45 (s+34.88) (s+26.83) (s+2.122) (s^2 + 1.17s + 0.5964)-----------------------------------------------------------------(s+47) (s+39) (s+26.34) (s^2 + 0.6618s + 0.5695) (s^2 + 2s + 100)
  • 29. Solutions to Problems 2-11Copyright © 2011 by John Wiley & Sons, Inc.ans =(b)ans =Ga symbolic3 2(s + 14) (s + 49 s + 62 s + 53)56 ----------------------------------------------------------2 2 3 2(s + 88 s + 33) (s + 56 s + 77) (s + 81 s + 76 s + 65)ans =Ga polynimialTransfer function:56 s^4 + 3528 s^3 + 41888 s^2 + 51576 s + 41552--------------------------------------------------------------------------------s^7 + 225 s^6 + 16778 s^5 + 427711 s^4 + 1.093e006 s^3 + 1.189e006 s^2+ 753676 s + 165165ans =Ga factoredZero/pole/gain:56 (s+47.72) (s+14) (s^2 + 1.276s + 1.111)---------------------------------------------------------------------------(s+87.62) (s+80.06) (s+54.59) (s+1.411) (s+0.3766) (s^2 + 0.9391s + 0.8119)16.a. Writing the node equations,Vo − Vis+Vos+ Vo = 0. Solve forVoVi=1s + 2.b. Thevenizing,Using voltage division, Vo (s) =Vi (s)21s12+ s +1s. Thus,Vo (s)Vi (s)=12s2+ s + 2
  • 30. Chapter 2: Modeling in the Frequency Domain 2-12Copyright © 2011 by John Wiley & Sons, Inc.17.a.Writing mesh equations(2s+2)I1(s) –2 I2(s) = Vi(s)-2I1(s) + (2s+4)I2(s) = 0But from the second equation, I1(s) = (s+2)I2(s). Substituting this in the first equation yields,(2s+2)(s+2)I2(s) –2 I2(s) = Vi(s)orI2(s)/Vi(s) = 1/(2s2+ 4s + 2)But, VL(s) = sI2(s). Therefore, VL(s)/Vi(s) = s/(2s2+ 4s + 2).b.i1(t) i2(t)2222222
  • 31. Solutions to Problems 2-13Copyright © 2011 by John Wiley & Sons, Inc.1 212 1(4 ) ( ) (2 ) ( ) ( )1 1(2 ) ( ) (4 2 ) 0I s I s V ss sI s ss s+ − + =− + + + + =Solving for I2(s):2 224 2( )(2 1)0( )( )4 2 (2 1) 4 6 1(2 1) (2 4 1)sV ssssV ssI ss s s ss ss s ss s+− += =+ − + + +− + + +Therefore,222( ) 2 ( ) 2( ) ( ) 4 6 1LV s sI s sV s V s s s= =+ +18.a.Writing mesh equations,(2s + 1)I1(s) – I2(s) = Vi(s)-I1(s) + (3s + 1 + 2/s)I2(s) = 0Solving for I2(s),
  • 32. Chapter 2: Modeling in the Frequency Domain 2-14Copyright © 2011 by John Wiley & Sons, Inc.I2 (s) =2s +1 Vi (s)−1 02s + 1 −1−13s2+ s + 2sSolving for I2(s)/Vi(s),I2 (s)Vi(s)=s6s3+ 5s2+ 4s + 2But Vo(s) = I2(s)3s. Therefore , G(s) = 3s2/(6s3+ 5s2+4s + 2).b. Transforming the network yields,Writing the loop equations,(s +ss2+1)I1(s) −ss2+1I2 (s) − sI3(s) = Vi(s)−ss2+1I1(s) + (ss2+1+1 +1s)I2 (s) − I3 (s) = 0−sI1(s) − I2 (s) +(2s +1)I3 (s) = 0Solving for I2(s),I2 (s) =s(s2+ 2s + 2)s4+ 2s3+ 3s2+ 3s + 2Vi(s)But, Vo(s) =I2(s)s =(s2+ 2s + 2)s4+ 2s3+ 3s2+ 3s + 2Vi(s). Therefore,Vo (s)Vi(s)=s2+ 2s + 2s4+ 2s3+ 3s2+ 3s + 219.a. Writing the nodal equations yields,
  • 33. Solutions to Problems 2-15Copyright © 2011 by John Wiley & Sons, Inc.VR(s) −Vi (s)2s+VR(s)1+VR (s) − VC (s)3s= 0−13sVR(s) +12s +13s⎛⎝⎞⎠VC (s) = 0Rewriting and simplifying,6s + 56sVR(s) −13sVC (s) =12sVi (s)−13sVR(s) +3s2+ 26s⎛⎝⎜ ⎞⎠VC (s) = 0Solving for VR(s) and VC(s),VR(s) =12sVi (s) −13s03s2+ 26s6s + 56s−13s−13s3s2+ 26s; VC (s) =6s + 56s12sVi (s)−13s06s + 56s−13s−13s3s2+ 26sSolving for Vo(s)/Vi(s)Vo (s)Vi (s)=VR(s) − VC (s)Vi (s)=3s26s3+ 5s2+ 4s + 2b. Writing the nodal equations yields,(V1(s) − Vi (s))s+(s2+1)sV1(s) +(V1(s) − Vo (s)) = 0(Vo (s) − V1(s)) + sVo(s) +(Vo (s) − Vi (s))s= 0Rewriting and simplifying,(s +2s+1)V1(s) − Vo (s) =1sVi (s)V1(s) + (s +1s+ 1)Vo (s) =1sVi (s)
  • 34. Chapter 2: Modeling in the Frequency Domain 2-16Copyright © 2011 by John Wiley & Sons, Inc.Solving for Vo(s)Vo(s) =(s2+ 2s + 2)s4+ 2s3+ 3s2+ 3s + 2Vi (s).Hence,Vo (s)Vi (s)=(s2+ 2s + 2)s4+ 2s3+ 3s2+ 3s + 220.a.Mesh:(4+4s)I1(s) - (2+4s)I2(s) - 2I3(s) = V(s)- (2+4s)I1(s) + (14+10s)I2(s) - (4+6s)I3(s) = 0-2I1(s) - (4+6s)I2(s) + (6+6s+ 9s)I3(s) = 0Nodal:11 1 ( ( ) ( ))( ( ) ( )) ( )02 2 4 4 6oV s V sV s V s V ss s−−+ + =+ +1( ( ) ( )) ( ) ( ( ) ( ))04 6 8 9/o o oV s V s V s V s V ss s− −+ + =+or1/92 4 6248
  • 35. Solutions to Problems 2-17Copyright © 2011 by John Wiley & Sons, Inc.2126s + 12s + 5 1 1( ) ( ) ( )12s 14 4 6 4 2oV s V s V ss s⎡ ⎤ ⎡ ⎤− =⎢ ⎥ ⎢ ⎥+ + +⎣ ⎦⎣ ⎦211 24s + 43s + 54( ) ( ) ( )6 4 216 144 9osV s V s V ss s⎡ ⎤⎡ ⎤− + =⎢ ⎥⎢ ⎥+ +⎣ ⎦ ⎣ ⎦b.Program:syms s V %Construct symbolic object for frequency%variable s and V.Mesh EquationsA2=[(4+4*s) V -2-(2+4*s) 0 -(4+6*s)-2 0 (6+6*s+(9/s))] %Form Ak = A2.A=[(4+4*s) -(2+4*s) -2-(2+4*s) (14+10*s) -(4+6*s)-2 -(4+6*s) (6+6*s+(9/s))] %Form A.I2=det(A2)/det(A); %Use Cramers Rule to solve for I2.Gi=I2/V; %Form transfer function, Gi(s) = I2(s)/V(s).G=8*Gi; %Form transfer function, G(s) = 8*I2(s)/V(s).G=collect(G); %Simplify G(s).G(s) via Mesh Equations %Display label.pretty(G) %Pretty print G(s)Nodal EquationsA2=[(6*s^2+12*s+5)/(12*s^2+14*s+4) V/2-1/(6*s+4) s*(V/9)] %Form Ak = A2.A=[(6*s^2+12*s+5)/(12*s^2+14*s+4) -1/(6*s+4)-1/(6*s+4) (24*s^2+43*s+54)/(216*s+144)]%Form A.Vo=simple(det(A2))/simple(det(A));%Use Cramers Rule to solve for Vo.G1=Vo/V; %Form transfer function, G1(s) = Vo(s)/V(s).G1=collect(G1); %Simplify G1(s).G(s) via Nodal Equations %Display label.pretty(G1) %Pretty print G1(s)Computer response:
  • 36. Chapter 2: Modeling in the Frequency Domain 2-18Copyright © 2011 by John Wiley & Sons, Inc.ans =Mesh EquationsA2 =[ 4*s + 4, V, -2][ - 4*s - 2, 0, - 6*s - 4][ -2, 0, 6*s + 9/s + 6]A =[ 4*s + 4, - 4*s - 2, -2][ - 4*s - 2, 10*s + 14, - 6*s - 4][ -2, - 6*s - 4, 6*s + 9/s + 6]ans =G(s) via Mesh Equations3 248 s + 96 s + 112 s + 36----------------------------3 248 s + 150 s + 220 s + 117ans =Nodal EquationsA2 =[ (6*s^2 + 12*s + 5)/(12*s^2 + 14*s + 4), V/2][ -1/(6*s + 4), (V*s)/9]A =[ (6*s^2 + 12*s + 5)/(12*s^2 + 14*s + 4), -1/(6*s + 4)][ -1/(6*s + 4), (24*s^2 + 43*s + 54)/(216*s + 144)]ans =G(s) via Nodal Equations3 248 s + 96 s + 112 s + 36----------------------------3 248 s + 150 s + 220 s + 117
  • 37. Solutions to Problems 2-19Copyright © 2011 by John Wiley & Sons, Inc.21.a.51 652 61( ) 5 102 101( ) 102 10Z s xx sZ sx s−−= += +Therefore,( )( )215( ) 1( ) 5 1sZ sZ s s+− = −+b.5 515 ( 5)( ) 10 1 10sZ ss s+⎛ ⎞= + =⎜ ⎟⎝ ⎠( )5 525 ( 10)( ) 10 1 105 5sZ ss s+⎛ ⎞= + =⎜ ⎟+ +⎝ ⎠Therefore,( )( )22110( )( ) 5s sZ sZ s s+− = −+22.a.51 652 61( ) 4 104 101( ) 1.1 104 10Z s xx sZ s xx s−−= += +Therefore,1 21( ) ( ) ( 0.98)( ) 1.275( ) ( 0.625)Z s Z s sG sZ s s+ += =+b.1151 65952 6310( ) 4 100.25 104 101027.5( ) 6 100.25 10110 10sZ s xxxssZ s xxxs= ++= ++Therefore,21 221( ) ( ) 2640 8420 4275( ) 1056 3500 2500Z s Z s s sZ s s s+ + +=+ +
  • 38. Chapter 2: Modeling in the Frequency Domain 2-20Copyright © 2011 by John Wiley & Sons, Inc.23.Writing the equations of motion, where x2(t) is the displacement of the right member of springr,(5s2+4s+5)X1(s) -5X2(s) = 0-5X1(s) +5X2(s) = F(s)Adding the equations,(5s2+4s)X1(s) = F(s)From which, 1X (s) 1 1/5F(s) s(5s 4) s(s 4/5)= =+ +.24.Writing the equations of motion,(s2+ s +1)X1 (s) − (s +1)X2 (s) = F(s)−(s +1)X1(s) + (s2+ s +1)X2(s) = 0Solving for X2(s),X2 (s) =(s2+ s + 1) F(s)−(s +1) 0⎡⎣⎢⎢⎤⎦⎥⎥(s2+ s +1) −(s +1)−(s +1) (s2+ s +1)⎡⎣⎢⎢⎤⎦⎥⎥=(s +1)F(s)s2(s2+ 2s + 2)From which,X2 (s)F(s)=(s +1)s2(s2+ 2s + 2).25.Let X1(s) be the displacement of the left member of the spring and X3(s) be the displacement of themass.Writing the equations of motion2x1(s) − 2x2 (s) = F(s)−2X1(s) + (5s + 2)X2(s) − 5sX3(s) = 0−5sX2 (s) + (10s2+ 7s)X3(s) = 0Solving for X2(s),X2(s) =⎪⎪⎪5s2+10 -10-1051s+10⎪⎪⎪⎪⎪⎪5s2+10 F(s)-10 0⎪⎪⎪=s(s2+50s+2)10F(s)
  • 39. Solutions to Problems 2-21Copyright © 2011 by John Wiley & Sons, Inc.Thus,X2 (s)F(s)=110(10s + 7)s(5s +1)26.21 221 2( 6 9) ( ) (3 5) ( ) 0(3 5) ( ) (2 5 5) ( ) ( )s s X s s X ss X s s s X s F s+ + − + =− + + + + =Solving for X1(s);21 4 3 2220 (3 5)( ) (2 5 5) (3 5) ( )( )2 17 44 45 20( 6 9) (3 5)(3 5) (2 5 5)sF s s s s F sX ss s s ss s ss s s− +⎡ ⎤⎢ ⎥+ + +⎣ ⎦= =+ + + +⎡ ⎤+ + − +⎢ ⎥− + + +⎣ ⎦Thus G(s) = X1(s)/F(s) = 4 3 2(3 5)2 17 44 45 20ss s s s++ + + +27.Writing the equations of motion,21 221 2 322 3(4 2 6) ( ) 2 ( ) 02 ( ) (4 4 6) ( ) 6 ( ) ( )6 ( ) (4 2 6) ( ) 0s s X s sX ssX s s s X s X s F sX s s s X s+ + − =− + + + − =− + + + =Solving for X3(s),223 3 2222(4 2 6) 2 02 (4 4 6) ( )0 6 0 3 ( )( )(8 12 26 18)(4 2 6) 2 02 (4 4 6) 60 6 (4 2 6)s s ss s s F sF sX ss s s ss s ss s ss s+ + −− + +−= =+ + ++ + −− + + −− + +From which, 33 2( ) 3( ) (8 12 26 18)X sF s s s s s=+ + +.28.a.21 2 321 2 31 2 3(4s 8s 5)X (s) 8sX (s) 5X (s) F(s)8sX (s) (4s 16s)X (s) 4sX (s) 05X (s) 4sX (s) (4s 5)X (s) 0+ + − − =− + + − =− − + + =Solving for X3(s),
  • 40. Chapter 2: Modeling in the Frequency Domain 2-22Copyright © 2011 by John Wiley & Sons, Inc.22 23(4s 8s 5) -8s F(s)8s (4s 16s) 0 8s (4s 16s)F(s)5 -4s 0 5 4X (s)s+ +− + − +− − −= =Δ Δor,33 2X (s) 13s 20F(s) 4s(4s 25s 43s 15)+=+ + +b.21 2 321 2 31 2 3(8s 4s 16)X (s) (4s 1)X (s) 15X (s) 0(4s 1)X (s) (3s 20s 1)X (s) 16sX (s) F(s)15X (s) 16sX (s) (16s 15)X (s) 0+ + − + − =− + + + + − =− − + + =Solving for X3(s),22 23(8s 4s 16) -(4s+1) 0(4s+1) (3s 20s+1) F(s) (8s 4s 16) -(4s+1)-F(s)15 -16s 0 15 16X (s)s+ +− + + +− − −= =Δ ΔorX3(s)F(s) =3 25 4 3 2128 64 316 15384 1064 3476 165s s ss s s s+ + ++ + +29.Writing the equations of motion,21 2 321 2 321 2(4 4 8) ( ) 4 ( ) 2 ( ) 04 ( ) (5 3 4) ( ) 3 ( ) ( )2 ( ) 3 ( ) (5 5 5) 0s s X s X s sX sX s s s X s sX s F ssX s sX s s s+ + − − =− + + + − =− − + + + =
  • 41. Solutions to Problems 2-23Copyright © 2011 by John Wiley & Sons, Inc.30.a.Writing the equations of motion,21 221 2(5 9 9) ( ) ( 9) ( ) 0( 9) ( ) (3 12) ( ) ( )s s s s ss s s s s T sθ θθ θ+ + − + =− + + + + =b.Definingθ1(s) = rotation of J1θ2 (s) = rotation between K1 and D1θ3(s) = rotation of J3θ4 (s) = rotation of right - hand side of K2the equations of motion are(J1s2+ K1)θ1(s) − K1θ2 (s) = T(s)−K1θ1(s) + (D1s + K1 )θ2 (s) − D1sθ3(s) = 0−D1sθ2 (s) + (J2s2+ D1s + K2 )θ3(s) − K2θ4(s) = 0−K2θ3(s) + (D2s + (K2 + K3))θ4 (s) = 031.Writing the equations of motion,(s2+ 2s +1)θ1 (s) − (s +1)θ2 (s) = T(s)−(s +1)θ1(s) + (2s +1)θ2(s) = 0Solving for θ2 (s)θ2 (s) =(s2+ 2s +1) T(s)−(s +1) 0(s2+ 2s +1) −(s +1)−(s +1) (2s +1)=T(s)2s(s +1)Hence,θ2 (s)T(s)=12s(s + 1)32.Reflecting impedances to θ3,
  • 42. Chapter 2: Modeling in the Frequency Domain 2-24Copyright © 2011 by John Wiley & Sons, Inc.(Jeqs2+Deqs)θ3(s) = T(s) (N4 N2N3N1)Thus,θ3(s)T(s)=N4N2N3 N1Jeq s2+ DeqswhereJeq = J4+J5+(J2+J3)N4N3⎛⎝⎜⎜⎞⎠⎟⎟2+ J1N4 N2N3 N1⎛⎝⎜⎜⎞⎠⎟⎟2, andDeq = (D4 + D5 ) + (D2 + D3)(N4N3)2+ D1(N4N2N3N1)233.Reflecting all impedances to θ2(s),{[J2+J1(N2N1)2+J3 (N3N4 )2]s2 + [f2+f1(N2N1)2+f3(N3N4)2]s + [K(N3N4)2]}θ2(s) = T(s)N2N1Substituting values,{[1+2(3)2+16(14 )2]s2 + [2+1(3)2+32(14 )2]s + 64(14 )2}θ2(s) = T(s)(3)Thus,θ2(s)T(s) =320s2+13s+434.Reflecting impedances to θ2,200 + 3505⎛⎝⎜⎞⎠⎟2+ 200525x505⎛⎝⎜⎞⎠⎟2⎡⎣⎢⎢⎤⎦⎥⎥s2+ 1000525x505⎛⎝⎜⎞⎠⎟2⎡⎣⎢⎢⎤⎦⎥⎥s + 250 + 3505⎛⎝⎜⎞⎠⎟2⎡⎣⎢⎢⎤⎦⎥⎥=505⎛⎝⎜⎞⎠⎟T(s)Thus,θ2 (s)T(s)=101300s2+ 4000s + 55035.Reflecting impedances and applied torque to respective sides of the spring yields the followingequivalent circuit:
  • 43. Solutions to Problems 2-25Copyright © 2011 by John Wiley & Sons, Inc.Writing the equations of motion,2θ2(s) -2 θ3(s) = 4.231T(s)-2θ2(s) + (0.955s+2)θ3(s) = 0Solving for θ3(s),( )32 4.231 ( )2 0 8.462 ( ) 4.43 ( )( )2 2 1.912 0.955 2T sT s T sss ssθ−= = =−− +Hence, 3 ( ) 4.43( )sT s sθ= . But, 4 3( ) 0.192 ( )s sθ θ= . Thus, 4 ( ) 0.851( )sT s sθ= .36.Reflecting impedances and applied torque to respective sides of the viscous damper yields thefollowingequivalent circuit:Writing the equations of motion,22 32 3 43 4( 2 ) ( ) 2 ( ) 3 ( )2 ( ) (2 3) ( ) 3 ( ) 03 ( ) ( 3) ( ) 0s s s s s T ss s s s ss s sθ θθ θ θθ θ+ − =− + + − =− + + =Solving for θ4 (s) ,2 0.955
  • 44. Chapter 2: Modeling in the Frequency Domain 2-26Copyright © 2011 by John Wiley & Sons, Inc.4 2( 2) 2 3 ( )2 (2 3) 00 3 0 18 ( )( )( 2) 2 0 (2 9 6)2 (2 3) 30 3 ( 3)s s s T ss sT sss s s s s ss ssθ+ −− +−= =+ − + +− + −− +But, θL (s) = 5θ4 (s) . Hence,42( ) 90( ) (2 9 6)sT s s s sθ=+ +37.Reflect all impedances on the right to the viscous damper and reflect all impedances and torques on theleft to the spring and obtain the following equivalent circuit:Writing the equations of motion,(J1eqs2+K)θ2(s) -Kθ3(s) = Teq(s)-Kθ2(s)+(Ds+K)θ3(s) -Dsθ4(s) = 0-Dsθ3(s) +[J2eqs2 +(D+Deq)s]θ4(s) = 0where: J1eq = J2+(Ja+J1)(N2N1)2; J2eq = J3+(JL+J4)(N3N4)2; Deq = DL(N3N4)2; θ2(s) = θ1(s)N1N2.
  • 45. Solutions to Problems 2-27Copyright © 2011 by John Wiley & Sons, Inc.38.Reflect impedances to the left of J5 to J5 and obtain the following equivalent circuit:Writing the equations of motion,[Jeqs2+(Deq+D)s+(K2+Keq)]θ5(s) -[Ds+K2]θ6(s) = 0-[K2+Ds]θ5(s) + [J6s2+2Ds+K2]θ6(s) = T(s)From the first equation,θ6(s)θ5(s)=Jeqs2+(Deq+D)s+ (K2+Keq)Ds+K2. But,θ5(s)θ1(s)=N1N3N2N4. Therefore,θ6(s)θ1(s)=N1N3N2N4 ⎝⎜⎛⎠⎟⎞Jeqs2+(Deq+D)s+ (K2+Keq)Ds+K2,where Jeq = [J1(N4N2N3N1)2+ (J2+J3)(N4N3)2+ (J4+J5)], Keq = K1(N4N3)2, andDeq = D[(N4N2N3N1)2+ (N4N3)2+ 1].39.Draw the freebody diagrams,
  • 46. Chapter 2: Modeling in the Frequency Domain 2-28Copyright © 2011 by John Wiley & Sons, Inc.Write the equations of motion from the translational and rotational freebody diagrams,(Ms2+2fv s+K2)X(s) -fvrsθ(s) = F(s)-fvrsX(s) +(Js2+fvr2s)θ(s) = 0Solve for θ(s),θ(s) =Ms2+2fvs+K2F(s)-fvrs 0Ms2+2fvs+K2-fvrs-fvrs Js2+fvr2s=fvrF(s)JMs3+(2Jfv+Mfvr2)s2+(JK2+fv2r2)s+K2fvr2From which,θ(s)F(s) =fvrJMs3+(2Jfv+Mfvr2)s2+(JK2+fv2r2)s+K2fvr2 .40.Draw a freebody diagram of the translational system and the rotating member connected to thetranslational system.From the freebody diagram of the mass, F(s) = (2s2+2s+3)X(s). Summing torques on the rotatingmember,(Jeqs2 +Deqs)θ(s) + F(s)2 = Teq(s). Substituting F(s) above, (Jeqs2 +Deqs)θ(s) + (4s2+4s+6)X(s) =Teq(s). However, θ(s) =X(s)2 . Substituting and simplifying,Teq = [(Jeq2 +4)s2 +(Deq2 +4)s+6]X(s)But, Jeq = 3+3(4)2 = 51, Deq = 1(2)2 +1 = 5, and Teq(s) = 4T(s). Therefore,22 3
  • 47. Solutions to Problems 2-29Copyright © 2011 by John Wiley & Sons, Inc.[592s2 +132s+6]X(s) = 4T(s). Finally,X(s)T(s) = 2859 13 12s s+ +.41.Writing the equations of motion,(J1s2+K1)θ1(s) - K1θ2(s) = T(s)-K1θ1(s) + (J2s2+D3s+K1)θ2(s) +F(s)r -D3sθ3(s) = 0-D3sθ2(s) + (J2s2+D3s)θ3(s) = 0where F(s) is the opposing force on J2 due to the translational member and r is the radius of J2. But,for the translational member,F(s) = (Ms2+fvs+K2)X(s) = (Ms2+fvs+K2)rθ(s)Substituting F(s) back into the second equation of motion,(J1s2+K1)θ1(s) - K1θ2(s) = T(s)-K1θ1(s) + [(J2 + Mr2)s2+(D3 + fvr2)s+(K1 + K2r2)]θ2(s) -D3sθ3(s) = 0-D3sθ2(s) + (J2s2+D3s)θ3(s) = 0Notice that the translational components were reflected as equivalent rotational components by thesquare of the radius. Solving for θ2(s), θ2 (s) =K1(J3s2+ D3s)T(s)Δ, where Δ is thedeterminant formed from the coefficients of the three equations of motion. Hence,θ2 (s)T(s)=K1(J3s2+ D3s)ΔSinceX(s) = rθ2 (s),X(s)T(s)=rK1(J3s2+ D3s)Δ42.KtRa=TstallEa=10050= 2 ; Kb =Eaωno− load=50150=13Also,Jm = 5+18(13)2= 7; Dm = 8+36(13)2= 12.Thus,θm (s)Ea (s)=2/ 71 2( (12 ))7 3s s + +=2/ 738( )21s s +Since θL(s) =13θm(s),
  • 48. Chapter 2: Modeling in the Frequency Domain 2-30Copyright © 2011 by John Wiley & Sons, Inc.θL (s)Ea (s)=22138( )21s s +.43.The parameters are:KtRa=TsEa=55= 1; Kb =Eaω=5600π2π160=14; Jm =1614⎛⎝⎞⎠2+ 412⎛⎝⎞⎠2+1 = 3; Dm = 3214⎛⎝⎞⎠2= 2Thus,θm (s)Ea (s)=13s(s +13(2 + (1)(14)))=13s(s + 0.75)Since θ2(s) =14θm(s),θ2 (s)Ea (s)=112s(s + 0.75).44.The following torque-speed curve can be drawn from the data given:vT10050500 1000Therefore,KtRa=TstallEa=10012; Kb =Eaωno− load=121333.33. Also, Jm = 7+105(16)2= 9.92; Dm =3. Thus,θm (s)Ea (s)=100 112 9.921( (3.075))9.92s s⎛ ⎞⎜ ⎟⎝ ⎠+=0.84( 0.31)s s +. Since θL(s) =16θm(s),θL (s)Ea (s)=0.14( 0.31)s s +.55600 1333.33
  • 49. Solutions to Problems 2-31Copyright © 2011 by John Wiley & Sons, Inc.45.From Eqs. (2.45) and (2.46),RaIa(s) + Kbsθ(s) = Ea(s) (1)Also,Tm(s) = KtIa(s) = (Jms2+Dms)θ(s). Solving for θ(s) and substituting into Eq. (1), and simplifyingyieldsIa (s)Ea (s)=1Ra(s +DmJm)s +Ra Dm + KbKtRaJm(2)Using Tm(s) = KtIa(s) in Eq. (2),Tm(s)Ea (s)=KtRa(s +DmJm)s +Ra Dm + KbKtRaJm46.For the rotating load, assuming all inertia and damping has been reflected to the load,(JeqLs2+DeqLs)θL(s) + F(s)r = Teq(s), where F(s) is the force from the translational system, r=2 isthe radius of the rotational member, JeqL is the equivalent inertia at the load of the rotational load andthe armature, and DeqL is the equivalent damping at the load of the rotational load and the armature.Since JeqL = 1(2)2 +1 = 5, and DeqL = 1(2)2 +1 = 5, the equation of motion becomes, (5s2+5s)θL(s)+ F(s)r = Teq(s). For the translational system, (s2+s)X(s) = F(s). Since X(s) = 2θL(s), F(s) =(s2+s)2θL(s). Substituting F(s) into the rotational equation, (9s2+9s) θL(s) = Teq(s). Thus, theequivalent inertia at the load is 9, and the equivalent damping at the load is 9. Reflecting these backto the armature, yields an equivalent inertia of94 and an equivalent damping of94 . Finally,KtRa= 1;Kb = 1. Hence,θm(s)Ea(s) =49s(s+49(94+1))=49s(s+139 ). Since θL(s) =12 θm(s),θL(s)Ea(s) =29s(s+139 ). ButX(s) = rθL(s) = 2θL(s). therefore,X(s)Ea(s)=49s(s+139 ).
  • 50. Chapter 2: Modeling in the Frequency Domain 2-32Copyright © 2011 by John Wiley & Sons, Inc.47.The equations of motion in terms of velocity are:[M1s +( fv1 + fv3) +K1s+K2s]V1(s) −K2sV2(s) − fv3V3(s) = 0−K2sV1(s) + [M2s + ( fv2 + fv4) +K2s]V2 (s) − fv4V3 (s) = F(s)− fv3V1 (s) − fv4V2 (s) +[M3s + fV3 + fv4]V3(S) = 0For the series analogy, treating the equations of motion as mesh equations yieldsIn the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.For the parallel analogy, treating the equations of motion as nodal equations yieldsIn the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.48.Writing the equations of motion in terms of angular velocity, Ω(s) yields
  • 51. Solutions to Problems 2-33Copyright © 2011 by John Wiley & Sons, Inc.(J1s + D1 +K1s)Ω1(s) − (D1 +K1s)Ω2 (s) = T(s)−(D1 +K1s)Ω1(s) + (J2s + D1 +(K1 + K2 )s)Ω2 (s) = 0−K2sΩ2(s) − D2Ω3(s) + (D2 +K2s)Ω4 (s) = 0(J3s + D2 +K3s)Ω3(s) − D2Ω4 (s) = 0For the series analogy, treating the equations of motion as mesh equations yieldsIn the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.For the parallel analogy, treating the equations of motion as nodal equations yieldsIn the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.49.An input r1 yields c1 = 5r1+7. An input r2 yields c2 = 5r2 +7. An input r1 +r2 yields, 5(r1+r2)+7 =5r1+7+5r2 = c1+c2-7. Therefore, not additive. What about homogeneity? An input of Kr1 yields c =5Kr1+7 ≠ Kc1. Therefore, not homogeneous. The system is not linear.50.a. Let x = δx+0. Therefore,
  • 52. Chapter 2: Modeling in the Frequency Domain 2-34Copyright © 2011 by John Wiley & Sons, Inc.δx..+3δx.+2δx =sin (0+δx)But, sin (0+δx) = sin 0 +d sinxdx ⎮x=0δx = 0+cosx ⎮x=0δx = δxTherefore, δx..+3δx.+2δx = δx. Collecting terms, δx..+3δx.+δx = 0 .b. Let x = δx+π. Therefore,δx..+3δx.+2δx =sin (π+δx)But, sin (π+δx) = sin π +d sinxdx ⎮x=πδx = 0+cosx ⎮x=πδx = −δxTherefore, δx..+3δx.+2δx = -δx. Collecting terms, δx..+3δx.+3δx = 0 .51.If x = 0 + δx,δx...+ 10δx..+ 31δx.+ 30δx = e-(δx)But e-(δx)= e-0+de-xdx ⎮x=0δx = 1 - e-x⎮x=0δx = 1 - δxTherefore, δx...+ 10δx..+ 31δx.+ 30δx =1 - δx, or, δx...+ 10δx..+ 31δx.+ 31δx =1.52.The given curve can be described as follows:f(x) = -6 ; -∞<x<-3;f(x) = 2x; -3<x<3;f(x) = 6; 3<x<+∞Thus,a. 17 50 6b. 17 50 2 or 17 48 0c. 17 50 6x x xx x x x x x xx x x+ + = −+ + = + + =+ + =&& &&& & && &&& &53.The relationship between the nonlinear spring’s displacement, xs(t) and its force, fs(t) isxs (t) = 1 − e− fs (t)Solving for the force,fs (t) = −ln(1 − xs (t)) (1)Writing the differential equation for the system by summing forces,d2x(t)dt2 +dx(t)dt− ln(1− x(t)) = f (t) (2)
  • 53. Solutions to Problems 2-35Copyright © 2011 by John Wiley & Sons, Inc.Letting x(t) = x0 + δx and f(t) = 1 + δf, linearize ln(1 – x(t)).ln(1− x) − ln(1 − x0 ) =d ln(1 − x)dx x =x0δxSolving for ln(1 – x),ln(1− x) = ln(1 − x0 ) −11 − x x= x0δx = ln(1− x0) −11− x0δx (3)When f = 1, δx = 0. Thus from Eq. (1), 1 = -ln(1 – x0 ). Solving for x0,1 – x0 = e-1, or x0 = 0.6321.Substituting x0 = 0.6321 into Eq. (3),ln(1- x) = ln(1 – 0.6321) -11- 0.6321δx = -1 - 2.718δxPlacing this value into Eq. (2) along with x(t) = x0 + δx and f(t) = 1 + δf, yields the linearizeddifferential equation,d2δxdt2 +dδxdt+ 1+ 2.718δx = 1 +δford2δxdt2 +dδxdt+ 2.718δx = δfTaking the Laplace transform and rearranging yields the transfer function,δx(s)δf(s)=1s2+ s + 2.71854.First assume there are n plates without the top plate positioned at a displacement of y2(t) wherey2(t) = 0 is the position of the unstretched spring. Assume the system consists of mass M, where M isthe mass of the dispensing system and the n plates, viscous friction, fv, where the viscous frictionoriginates where the piston meets the sides of the cylinder, and of course the spring with springconstant, K. Now, draw the freebody diagram shown in Figure (b) where Wn is the total weight of then dishes and the piston. If we now consider the current position, y2(0),
  • 54. Chapter 2: Modeling in the Frequency Domain 2-36Copyright © 2011 by John Wiley & Sons, Inc.Restaurant Plate Dispenserthe equilibrium point and define a new displacement, y1(t), which is measured from equilibrium, wecan write the force in the spring as Ky2(t) = Ky2(0) + Ky1(t). Changing variables from y2(t) to y1(t),we sum forces and get,Md2y1dt2 + fvdy1dt + Ky1 + Ky2(0) + Wn = 0 (1)whered2y2dt2 =d2y1dt2 anddy2dt =dy1dt . But, Ky2(0) = -Wn , since it is the component of the springforce that balances the weight at equilibrium when y1 = 0. Thus, the differential equation becomes,Md2y1dt2 + fvdy1dt + Ky1 = 0 (2)When the top plate is added, the spring is further compressed by an amount,WDK , where WD is theweight of the single dish, and K is the spring constant. We can think of this displacement as an initialcondition. Thus, y1(0-) = -WDK anddy1dt (0-) =0, and y1(t) = 0 is the equilibrium position of thespring with n plates rather than the unstretched position. Taking the Laplace transform of equation(2), using the initial conditions,
  • 55. Solutions to Problems 2-37Copyright © 2011 by John Wiley & Sons, Inc.M(s2Y1(s) + sWDK ) + fv(sY1(s) +WDK ) + KY1(s) = 0 (3)or(Ms2 + fvs + K)Y1(s) = -WDK(Ms + fv ) (4)Now define a new position reference, Y(s), which is zero when the spring is compressed with theinitial condition,Y(s) = Y1(s) +WDKs (5)orY1(s) = Y(s) -WDKs (6)Substituting Y1(s) in Equation (4), we obtain,(Ms2 + fvs + K)Y(s) =WDs = F(s) (7)a differential equation that has an input and zero initial conditions. The schematic is shown in Figure(c). Forming the transfer function,Y(s)F(s) , we show the final result in Figure (d), where for theremoval of the top plate, F(s) is always a step, F(s) =WDs .55.We have ψφφφ &&&&& )(aJkbJ =++Assuming zero initial conditions and obtaining Laplace transform on both sides of the equation weobtain:)()()()()(2saJsksbssJs Ψ=Φ+Φ+Φ &&From which we get:kbsJsaJss++=ΨΦ2)()(&&56.a. We choose Laplace transforms to obtain a solution. After substitution of numerical values theequations become:)(7.0)(9.0)(tNtCdttdC+−=)()(02.0)(1.0)(tItNtCdttdN+−−=
  • 56. Chapter 2: Modeling in the Frequency Domain 2-38Copyright © 2011 by John Wiley & Sons, Inc.Obtaining Laplace transforms and substituting initial values we obtain:)(7.0)(9.047000500)( sNsCssC +−=−)()(02.0)(1.061100500)( sIsNsCssN +−−=−Both equations are manipulated as follows:9.0470005009.0)(7.0)(+++=sssNsC02.0)(02.06110050002.0)(1.0)(+++++−=ssIsssCsNSubstituting the first equation into the second one gets:)9.0)(02.0(088.092.002.0)(02.061100500)9.0)(02.0(4700050)9.0)(02.0(07.0102.0)(02.061100500)9.0)(02.0(4700050)( 2++++++++++−=+++++++++−=ssssssIsssssssIssssN088.092.04700050088.092.0)9.0(61100500)(088.092.09.0222++−+++++++=ssssssIsssFrom which we get the block diagram:088.092.0)9.0(2+++sss9.04700050+sb. LettingssI6106)(×= and after algebraic manipulations one gets:)1084.0)(8116.0(10545629040061100500)088.092.0(10545629040061100500)(52252++×++=++×++=sssssssssssN1084.0103.198116.0107101354.61084.08116.0447+×−+×−×=++++=ssssCsBsA
  • 57. Solutions to Problems 2-39Copyright © 2011 by John Wiley & Sons, Inc.Obtaining the inverse Laplace transform:tteetN 1084.048116.047103.19107101364.6)( −−×−×−×=57.a.‘Exact’:From Figure (a)( ) ( )0.0050.0051 1 313 18 8 8( ) ( )1 1 18 8 8sso ineeV s V sss s s s−− −−= = =⎛ ⎞+ + +⎜ ⎟⎝ ⎠Using Partial fraction expansions note that111)1(1+−=+ ssss. Thus, applying partial fractionexpansion to Vo(s) and taking the inverse Laplace transform yields,So1 1( 0.005)8 8( ) 3(1 ) ( ) 3(1 ) ( 0.005)t tov t e u t e u t− − −= − − − −‘Impulse’:1 10.0018758 8( ) ( ) 0.0151 1 0.1258 8o inV s V sss s= = =++ +In this case0.125( ) 0.001875 tov t e−=b.The following M-File will simulate both inputs:syms ss = tf(s);G=(1/8)/(s+(1/8));t=0:1e-4:10;for i=1:max(size(t)),if(i*1e-4 <= 5e-3)vinexact(i) = 3;elsevinexact(i) = 0;endendyexact = lsim(G,vinexact,t);yimpulse = 0.001875*exp(-(1/8)*t);plot(t,yexact,t,yimpulse)
  • 58. Chapter 2: Modeling in the Frequency Domain 2-40Copyright © 2011 by John Wiley & Sons, Inc.Resulting in the following figure:0 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.41.61.82x 10-3Both outputs are indistinguishable at this scale. However zooming closer to t=0 will showdifferences.58.a.At equilibrium022=dtHd. From which we get that 22HIkmg = ormgkIH 00 =b.Following the ‘hint’ procedure:IHHHHIIkHHHHHIIkdtHdm IHIH δδδδδδδδδδδδδ 0,0402000,04002022)()()(2)()()(2====+++−+++=After some algebraic manipulations this becomes:ImHkIHmHkIdtHdm δδδ20030202222−=Obtaining Laplace transform on both sides of the equation one obtains the transfer function:
  • 59. Solutions to Problems 2-41Copyright © 2011 by John Wiley & Sons, Inc.3020220022)()(mHkIsmHkIsIsH−−=δδ59.The two differential equations for this system are:0)()( =−+−+ wsawsasb xxCxxKxM &&&&0)()()( =−+−+−+ rxKxxCxxKxM wtswaswawus&&&&Obtaining Laplace transform on both sides gives0)()( 2=+−++ waasaab XsCKXKsCsMtsaawtaaus RKXsCKXKKsCsM =+−+++ )())(( 2Solving the first equation for sX and substituting into the second one gets( ) 2222)()()()()(sCKKsCsMKKsCsMKsCsMKsRXaaaabtaausaabtw+−+++++++=60.a. The three equations are transformed into the Laplace domain:SkCKkSSs S ψψ −=−~0)~( CKSkCs M−= ψCkPs 2=The three equations are algebraically manipulated to give:CksKkksSSSψψψ +++=~0MKksSkC ~ψψ+=CskP 2=By direct substitutions it is obtained that:
  • 60. Chapter 2: Modeling in the Frequency Domain 2-42Copyright © 2011 by John Wiley & Sons, Inc.022)~~()~1()~(SKKksKksKksSSMMM−++++=ψψψ022))~~()~1((SKKksKkskCSMM −+++=ψψψ0222))~~()~1((SKKksKksskkPSMM −+++=ψψψb.0)()(0==∞→ssSLimSs0)()(0==∞→ssCLimCs02022020)~~()~~()()( SKkkKkSkKKkSkkssPLimPSSSMs=−+=−==∞→ψψψψ61.Eliminate balT by direct substitution. This results in)()()()(022tTdttJtJtkJdtdJ dt+−−−= ∫θρθηθθ &Obtaining Laplace transform on both sides of this equation and eliminating terms one gets that:ρη +++=ΘkssssJsTd231)(62.a.We have thatφgmxm LLaL =&&LTLa xxx −=φLxL =From the second equationφφ gLvxxx TLTLa =−=−= &&&&&&&&&
  • 61. Solutions to Problems 2-43Copyright © 2011 by John Wiley & Sons, Inc.Obtaining Laplace transforms on both sides of the previous equationΦ=Φ− gLssVT from which )( 2LsgsVT +Φ=so that2022211)(ω+=+=+=ΦssLLgssLLsgssVTb.Under constant velocitysVsVT0)( = so the angle is2020 1)(ω+=ΦsLVsObtaining inverse Laplace transform)sin()( 000tLVt ωωφ = , the load will sway with a frequency 0ω .c. From φgmfxm LTTT −−&& and Laplace transformation we getTLTTLTLTTT XssLgmFVssLgmFsgmFsXsm 20222022 11)()(ωω +−=+−=Φ−=From which)(1))(()1(120222022020222022022 ωωωωωωasssmmsmsssLgmmsFXTLTLTTT++=+++=++=WhereTLmma += 1d. From part c)(1202202ωωasssmFsXFVTTTTT++==LetsFFT0= then202220222020)()(ωωωasDCssBsAasssmFsVTT++++=++=After partial fraction expansions, so)cos()( 0 θω +++= taCBtAtvTFrom which it is clear that ∞⎯⎯⎯ →⎯ ∞⎯→⎯tTv
  • 62. Chapter 2: Modeling in the Frequency Domain 2-44Copyright © 2011 by John Wiley & Sons, Inc.63.a. Obtaining Laplace transforms on both sides of the equation)()( 0 sKNNssN =− orKsNsN−= 0)(By inverse Laplace transformationKteNtN 0)( =b.Want to find the time at which00 2NeN Kt=Obtaining ln on both sides of the equationKt2ln=64.a.Converting each one of the impedances to its Laplace transform equivalent and applying thevoltage divider rule one getsZCsCsCZsCZSCZ1111+=+=)1()1(11111)(2CLZCLRsZCLRsLCZCsCsLRZCsCsPiPo++++=++++=
  • 63. Solutions to Problems 2-45Copyright © 2011 by John Wiley & Sons, Inc.SinceZPQ o=0 ,9.73550125.330236.0)1()1(1)( 22 ++=++++=ssCLZCLRsZCLRsLCZsPQiob. The steady state circuit becomesSo that60 102.330816341761 −=+=+= XZRPQ ic. Applying the final value theorem620102.319.73550125.330236.0)( −⎯→⎯=++=∞ Xssssq Limso65. The laplace transform of the systems output is2 2 2 2 2 22 2{ ( )} ( )4 ( ) 4ref ref refT T Ta f a fT t T ss s s f s s s fλπ πλ π λ π= = − + = ++ + + +£Dividing by the input one gets2 2 22( )4refT a f ssU s T s fλ πλ π= ++ +66.a. By direct differentiation )()()( )1(0 tVeeeVdttdV tetatααλαλααλ −−−==−b. αλαλ αeVeVLimtVLimVtett0)1(0)()( ===∞−−∞⎯→⎯∞⎯→⎯c.
  • 64. Chapter 2: Modeling in the Frequency Domain 2-46Copyright © 2011 by John Wiley & Sons, Inc.Lambda = 2.5;alpha = 0.1;V0=50;t=linspace(0,100);V=V0.*exp(Lambda.*(1-exp(-alpha.*t))/alpha);plot(t,V)gridxlabel(t (days))ylabel(mm^3 X 10^-3)0 10 20 30 40 50 60 70 80 90 10000.511.522.533.54x 1012t (days)mm3X10-3d. From the figure12105.3)( XV ≈∞ mm3X 10-3From part c121.05.20 106.350)( XeeVV ===∞ αλmm3X 10-367.Writing the equations of motion,(17.2s2+ 160s + 7000)Yf(s) – (130s + 7000)Yh(s) – 0Ycat(s) = Fup(s)- (130s+7000)Yf(s) + (9.1ss+ 130s + 89300)Yh(s) - 82300Ycat(s) = 0
  • 65. Solutions to Problems 2-47Copyright © 2011 by John Wiley & Sons, Inc.- 0Yf(s) - 82300Yh(s) + 1.6173 x106Ycat(s) = 0These equations are in the form AY=F, where det(A) = 2.5314 x 108(s2+ 15.47 s + 9283) (s2+8.119 s + 376.3)Using Cramer’s rule:Ycat (s)Fup(s)=0.04227(s + 53.85)(s2+15.47s + 9283)(s2+ 8.119s + 376.3)Yh(s)Fup(s)=0.8306(s + 53.85)(s2+15.47s + 9283)(s2+ 8.119s + 376.3)Yh(s) − Ycat (s)Fup(s)=0.7883(s + 53.85)(s2+ 15.47s + 9283)(s2+8.119s + 376.3)68.a.The first two equations are nonlinear because of the Tv products on their right hand side.Otherwise the equations are linear.b.To find the equilibria let 0*===dtdvdtdTdtdTLeading to0=−− νβTdTs0*=− TTv μβ0*=− cvkTThe first equilibrium is found by direct substitution. For the second equilibrium, solve the last twoequations for T*μβTvT =*andkcvT =*. Equating we get thatkcTβμ=Substituting the latter into the first equation after some algebraic manipulations we get thatβμdcksv −= . It follows thatβμ kcdskcvT −==*.69.a. FrommkFFamw•−= , we have: amkFFFamkFF mStLRmw O•••• ++++=+= (1)
  • 66. Chapter 2: Modeling in the Frequency Domain 2-48Copyright © 2011 by John Wiley & Sons, Inc.Substituting for the motive force, F, and the resistances FRo, FL, and Fst using the equations givenin the problem, yields the equation:amkvvACgmgmfvPF mhwwtot••⎟⎠⎞⎜⎝⎛••••••••••++++==25.0sincos ρααη(2)b. Noting that constant acceleration is assumed, the average values for speed and acceleration are:aav = 20 (km/h)/ 4 s = 5 km/h.s = 5x1000/3600 m/s2 = 1.389 m/s2 vav = 50 km/h = 50,000/3,600 m/s = 13.89 m/sThe motive force, F (in N), and power, P (in kW) can be found from eq. 2:Fav = 0.011 x 1590 x 9.8 + 0.5 x 1.2 x 0.3 x 2 x 13.892+ 1.2 x 1590 x 1.389 = 2891 NPav = Fav. v / η tot = 2891 x 13.89 / 0.9 = 44, 617 N.m/s = 44.62 kWTo maintain a speed of 60 km/h while climbing a hill with a gradient α = 5o, the car engine ormotor needs to overcome the climbing resistance:13585sin8.91590sin == ••••=oαgmFSt NThus, the additional power, Padd, the car needs after reaching 60 km/h to maintain its speed whileclimbing a hill with a gradient α = 5ois:η/vStFaddP •= = 1358 x 60 x 1000/(3,600 x 0.9) = 25, 149 W = 25.15 kWc. Substituting for the car parameters into equation 2 yields:dtdvvF /1590x1.22x0.3x1.2x0.59.8x1590x0.011 2++=or dtdvvtF /19080.364.171)( 2++= (3)To linearize this equation about vo = 50 km/h = 13.89 m/s, we use the truncated taylor series:
  • 67. Solutions to Problems 2-49Copyright © 2011 by John Wiley & Sons, Inc.)2)222(()(ooovvo vvvvvdvvdvvo−−≈− •==(4), from which we obtain:222289.1378.27 −=− ••= vvvvv oo (5)Substituting from equation (5) into (3) yields:dtdvvtF /190846.69014.171)( +−+= ordtdvvtFFFtFtF oRoe /19080146.694.171)()()( +=+−=+−= (6)Equation (6) may be represented by the following block-diagram:d. Taking the Laplace transform of the left and right-hand sides of equation (6) gives,(s)1908(s)01)( sVVsFe += (7)Thus the transfer function, Gv(s), relating car speed, V(s) to the excess motive force, Fe(s), when thecar travels on a level road at speeds around vo = 50 km/h = 13.89 m/s under windless conditions is:sFVsGv1908011(s)(s))(e +== (8)Car Speed,v(t)Gv+Fo = 69.46 N+MotiveForce,F (t)ExcessMotiveForce,Fe(t)FRo = 171.4 N
  • 68. Copyright © 2011 by John Wiley & Sons, Inc.T H R E EModelingin the Time DomainSOLUTIONS TO CASE STUDIES CHALLENGESAntenna Control: State-Space RepresentationFor the power amplifier,Ea(s)Vp(s) =150s+150 . Taking the inverse Laplace transform, ea.+150ea =150vp. Thus, the state equation isea•= −150ea +150vpFor the motor and load, define the state variables as x1 = θm and x2 = θ.m. Therefore,x.1 = x2 (1)Using the transfer function of the motor, cross multiplying, taking the inverse Laplace transform,and using the definitions for the state variables,x.2 = -1Jm(Dm+KtKaRa) x2 +KtRaJmea (2)Using the gear ratio, the output equation isy = 0.2x1 (3)Also, Jm = Ja+5(15 )2 = 0.05+0.2 = 0.25, Dm = Da+3(15 )2 = 0.01+0.12 = 0.13,KtRaJm=1(5)(0.25)= 0.8, and1Jm(Dm+KtKaRa) = 1.32. Using Eqs. (1), (2), and (3) along with the previous values, thestate and output equations are,x.=0 10 -1.32x +00.8ea ; y = 0.2 0 x
  • 69. 3-2 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.Aquifer: State-Space RepresentationC1dh1dt = qi1-qo1+q2-q1+q21 = qi1-0+G2(h2-h1)-G1h1+G21(H1-h1) =-(G2+G1+G21)h1+G2h2+qi1+G21H1C2dh2dt = qi2-q02+q3-q2+q32 = qi2-qo2+G3(h3-h2)-G2(h2-h1)+0 = G2h1-[G2+G3]h2+G3h3+qi2-qo2C3dh3dt = qi3-qo3+q4-q3+q43 = qi3-qo3+0-G3(h3-h2)+0 = G3h2-G3h3+qi3-qo3Dividing each equation by Ci and defining the state vector as x = [h1 h2 h3]Tx.=−(G1 + G2 + G3 )C1G2C10G2C2−(G2 + G3)C2G3C20G3C3−G3C3⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥x +qi1 + G21H1C1qi 2 − qo2C2qi3 − qo3C3⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥u(t)y =1 0 00 1 00 0 1⎡⎣⎢⎢⎤⎦⎥⎥xwhere u(t) = unit step function.ANSWERS TO REVIEW QUESTIONS1. (1) Can model systems other than linear, constant coefficients; (2) Used for digital simulation2. Yields qualitative insight3. That smallest set of variables that completely describe the system4. The value of the state variables5. The vector whose components are the state variables6. The n-dimensional space whose bases are the state variables7. State equations, an output equation, and an initial state vector (initial conditions)8. Eight9. Forms linear combinations of the state variables and the input to form the desired output10. No variable in the set can be written as a linear sum of the other variables in the set.11. (1) They must be linearly independent; (2) The number of state variables must agree with the order ofthe differential equation describing the system; (3) The degree of difficulty in obtaining the state equationsfor a given set of state variables.12. The variables that are being differentiated in each of the linearly independent energy storage elements
  • 70. 3-3 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.13. Yes, depending upon the choice of circuit variables and technique used to write the system equations.For example, a three -loop problem with three energy storage elements could yield three simultaneoussecond-order differential equations which would then be described by six, first-order differential equations.This exact situation arose when we wrote the differential equations for mechanical systems and thenproceeded to find the state equations.14. The state variables are successive derivatives.SOLUTIONS TO PROBLEMS1.Add the branch currents and node voltages to the network.Write the differential equation for each energy storage element.di2dt= v1di4dt= v2dvodt= i5Therefore, the state vector is x =i2i4vo⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥Now obtain v1, v2, and i5 in terms of the state variables. First find i1 in terms of the state variables.−vi + i1 + i3 + i5 + vo = 0But i3 = i1 − i2 and i5 = i3 − i4. Thus,−vi + i1 + (i1 − i2 ) + (i3 − i4 ) + vo = 0Making the substitution for i3 yields−vi + i1 + (i1 − i2 ) + ((i1 − i2 ) − i4 ) + vo = 0
  • 71. 3-4 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.Solving for i1i1 =23i2 +13i4 −13vo +13viThus,v1 = vi − i1 = −23i2 −13i4 +13vo +23viAlso,i3 = i1 − i2 = −13i2 +13i4 −13vo +13viandi5 = i3 − i4 = −13i2 −23i4 −13vo +13viFinally,v2 = i5 + vo = −13i2 −23i4 +23vo +13viUsing v1, v2, and i5, the state equation isx•=−23−1313−13−2323−13−23−13⎡⎣⎢⎢⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎥⎥x +231313⎡⎣⎢⎢⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎥⎥viy = 0 0 1[ ]x2.Add branch currents and node voltages to the schematic and obtain,Write the differential equation for each energy storage element.3233
  • 72. 3-5 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.1231312Ldvidtdivdt==Therefore the state vector is x =v1i3⎡⎣⎢⎢⎤⎦⎥⎥Now obtain vL and i2 , in terms of the state variables,1 2 1 1 3 1 1 33 3( 4 ) 11 3L Rv v v v i v i v v i= − = − = − + = − −2 1 3 1 3 1 31 1 1( )3 3 3i ii i i v v i v i v= − = − − = − − +Also, the output isy = iR = 4v1 + i3Hence,[ ]1 119 3911 32 24 1ivy•⎡ ⎤− − ⎡ ⎤⎢ ⎥⎢ ⎥= +⎢ ⎥⎢ ⎥⎢ ⎥− − ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦=x x0x3.Let C1 be the grounded capacitor and C 2 be the other. Now, writing the equations for the energystorage components yields,diLdt= vi − vC1dvC1dt= i1 − i2dvC2dt= i2 − i3(1)Thus the state vector is x =iLvC1vC2⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥. Now, find the three loop currents in terms of the state variablesand the input.Writing KVL around Loop 2 yields vC1= vC2+ i2 .Or,i2 = vC1− vC2
  • 73. 3-6 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.Writing KVL around the outer loop yields i3 + i2 = vi Or,i3 = vi − i2 = vi − vC1+ vC2Also, i1 − i3 = iL . Hence,i1 = iL + i3 = iL + vi − vC1+ vC2Substituting the loop currents in equations (1) yields the results in vector-matrix form,diLdtdvC1dtdvC2dt⎡⎣⎢⎢⎢⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎥⎥⎥=0 −1 01 −2 20 2 −2⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥iLvC1vC2⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥+11−1⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥viSince vo = i2 = vC1− vC2, the output equation isy = 0 1 1[ ]iLvC1vC2⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥4.Equations of motion in Laplace:21 2 321 2 321 2 3(2 3 2) ( ) ( 2) ( ) ( ) 0( 2) ( ) ( 2 2) ( ) ( ) ( )( ) ( ) ( 3 ) ( ) 0s s X s s X s sX ss X s s s X s sX s F ssX s sX s s s X s+ + − + − =− + + + + − =− − + + =Equations of motion in the time domain:231 1 21 22231 2 21 2223 31 222 3 2 2 02 2 2 ( )3 0dxd x dx dxx xdt dt dt dtdxdx d x dxx x f tdt dt dt dtd x dxdx dxdt dt dt dt+ + − − − =− − + + + − =− − + + =Define state variables:
  • 74. 3-7 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.1 1 1 11 12 23 2 2 3or (1)or (2)orz x x zdx dxz zdt dtz x x z= == == =2 24 45 3 3 53 36 6(3)or (4)or (5)ordx dxz zdt dtz x x zdx dxz zdt dt= == == = (6)Substituting Eq. (1) in (2), (3) in (4), and (5) in (6), we obtain, respectively:dz1dt= z2 (7)dz3dt= z4 (8)dz5dt= z6 (9)Substituting Eqs. (1) through (6) into the equations of motion in the time domain and solving for thederivatives of the state variables and using Eqs. (7) through (9) yields the state equations:1221 2 3 4 63441 2 3 4 65662 4 63 1 12 2 22 2 2 ( )3dzzdtdzz z z z zdtdzzdtdzz z z z z f tdtdzzdtdzz z zdt== − − + + +== + − − + +== + −The output is x3 = z5.In vector-matrix form:
  • 75. 3-8 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.[ ]0 1 0 0 0 0 01 1.5 1 0.5 0 0.5 00 0 0 1 0 0 0( )2 1 2 2 0 1 10 0 0 0 0 1 00 1 0 1 0 3 00 0 0 0 1 0f t•⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦=Z Zy Z5.Writing the equations of motion,21 2 321 2 321 2 3(2 2 1) ( ) ( ) ( 1) ( ) 0( ) ( 2 1) ( ) ( 1) ( ) 0( 1) ( ) ( 1) ( ) ( 2 2) ( ) ( )s s X s sX s s X ssX s s s X s s X ss X s s X s s s X s F s+ + − − + =− + + + − + =− + − + + + + =Taking the inverse Laplace transform,•• • • •1 1 1 2 3 3• •• • •1 2 2 2 3 3• • •• •1 1 2 2 3 3 32 2 02 02 2 ( )x x x x x xx x x x x xx x x x x x x f t+ + − − − =− + + + − − =− − − − + + + =Simplifying,•• • • •1 1 1 2 3 3•• • • •2 1 2 2 3 3•• • • •3 1 1 2 2 3 31 1 1 12 2 2 222 2 ( )x x x x x xx x x x x xx x x x x x x f t= − − + + += − − + += + + + − − +Defining the state variables,z1 = x1; z2 = x1•; z3 = x2 ; z4 = x2•; z5 = x3; z6 = x3•Writing the state equations using the simplified equations above yields,
  • 76. 3-9 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.• •1 1 2• ••2 1 2 1 4 6 5• •3 2 4• ••4 2 2 4 3 6 5• •5 3 6• ••6 3 2 1 4 3 6 51 1 1 12 2 2 222 2 ( )z x zz x z z z z zz x zz x z z z z zz x zz x z z z z z z f t= == = − − + + += == = − − + += == = + + + − − +Converting to vector-matrix form,[ ]•0 1 0 0 0 001 1 1 101 02 2 2 200 0 0 1 0 0 ( )00 1 1 2 1 100 0 0 0 0 111 1 1 1 2 21 0 0 0 0 0f ty⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥− − ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥= + ⎢ ⎥⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦− −⎢ ⎥⎣ ⎦=z zz6.Drawing the equivalent network,Writing the equations of motion,22 322 3(555.56 100) 100 3.33100 (100 100 100) 0s Ts sθ θθ θ+ − =− + + + =Taking the inverse Laplace transform and simplifying,555.563.33 T
  • 77. 3-10 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.2 2 32 3 3 30.18 0.18 0.0060Tθ θ θθ θ θ θ•••• •+ − =− + + + =Defining the state variables asx1 = θ2 , x2 =θ2•, x3 = θ3, x4 = θ3•Writing the state equations using the equations of motion and the definitions of the state variables1 22 2 2 3 1 33 44 3 2 3 3 1 3 42 10.18 0.18 0.006 0.18 0.18 0.0063.33 3.33x xx T x x Tx xx x x xy xθ θ θθ θ θ θθ•• •••• •• •== = − + + = − + +== = − − = − −= =,In vector-matrix form,[ ]0 1 0 0 00.18 0 0.18 0 0.0060 0 0 1 01 0 1 1 03.33 0 0 0Ty•⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦=x xx7.Drawing the equivalent circuit,Writing the equations of motion,10T(1/10)(102 )= 10 N-m/rad200(1/10)2=2 N-m/rad
  • 78. 3-11 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.2 32 3 43 412 ( ) 2 ( ) 10 ( )2 ( ) (3 2) ( ) 3 ( ) 03 ( ) 5 ( ) 0s s T ss s s s ss s s sθ θθ θ θθ θ− =− + + − =− + =Taking the inverse Laplace transform,2 312 ( ) 2 ( ) 10 ( )t t T tθ θ− = (1)−2θ2 (t) + 3θ3•(t) + 2θ3 −3θ4•(t) = 0 (2)−3θ3•(t) + 5θ4•(t) = 0 (3)From (3),θ3•(t) =53θ4•(t) and θ3(t) =53θ4 (t) (4)assuming zero initial conditions.From (1)2 3 41 5 5 5( ) ( ) ( ) ( ) ( )6 6 18 6t t T t t T tθ θ θ= + = + (5)Substituting (4) and (5) into (2) yields the state equation (notice there is only one equation),•4 425 5( ) ( ) ( )18 6t t T tθ θ= − +The output equation is given by,θL (t) =110θ4 (t)8.Solving Eqs. (3.44) and (3.45) in the text for the transfer functionsX1(s)F(s)andX2 (s)F(s):X 1 s0 K−F M 2 s2 K+M 1 s2 D s K+ + K−K− M 2 s2 K+= and X 2 sM 1 s2 D s K+ + 0K− FM 1 s2 D s K+ + K−K− M 2 s2 K+=Thus,X 1 sF sKM 2 M 1 s4D M 2 s3K M 2 s2K M 1 s2D K s+ + + +=andX 2 sF sM 1 s2D s K+ +M 2 M 1 s4D M 2 s3K M 2 s2K M 1 s2D K s+ + + +=Multiplying each of the above transfer functions by s to find velocity yields pole/zero cancellation atthe origin and a resulting transfer function that is third order.
  • 79. 3-12 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.9.a. . Using the standard form derived in the textbook,x•=0 1 0 00 0 1 00 0 0 1−100 −7 −10 −20⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥x +0001⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥r(t)c = 100 0 0 0[ ]xb. Using the standard form derived in the textbook,x•=0 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1−30 −1 −6 −9 −8⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥x +00001⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥r(t)c = 30 0 0 0 0[ ]x10.Program:anum=100;den=[1 20 10 7 100];G=tf(num,den)[Acc,Bcc,Ccc,Dcc]=tf2ss(num,den);Af=flipud(Acc);A=fliplr(Af)B=flipud(Bcc)C=fliplr(Ccc)bnum=30;den=[1 8 9 6 1 30];G=tf(num,den)[Acc,Bcc,Ccc,Dcc]=tf2ss(num,den);Af=flipud(Acc);A=fliplr(Af)B=flipud(Bcc)C=fliplr(Ccc)Computer response:ans =aTransfer function:100---------------------------------s^4 + 20 s^3 + 10 s^2 + 7 s + 100A =0 1 0 00 0 1 00 0 0 1-100 -7 -10 -20
  • 80. 3-13 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.B =0001C =100 0 0 0ans =bTransfer function:30------------------------------------s^5 + 8 s^4 + 9 s^3 + 6 s^2 + s + 30A =0 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1-30 -1 -6 -9 -8B =00001C =30 0 0 0 011.a. Using the standard form derived in the textbook,[ ]•0 1 0 0 00 0 1 0 0( )0 0 0 1 013 5 1 5 110 8 0 0 0r tc⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ ⎣ ⎦=x xx
  • 81. 3-14 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.b. Using the standard form derived in the textbook,[ ]•0 1 0 0 0 00 0 1 0 0 0( )0 0 0 1 0 00 0 0 0 1 00 0 8 13 9 16 7 12 2 1r tc⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦=x xx12.Program:anum=[8 10];den=[1 5 1 5 13]G=tf(num,den)[Acc,Bcc,Ccc,Dcc]=tf2ss(num,den);Af=flipud(Acc);A=fliplr(Af)B=flipud(Bcc)C=fliplr(Ccc)bnum=[1 2 12 7 6];den=[1 9 13 8 0 0]G=tf(num,den)[Acc,Bcc,Ccc,Dcc]=tf2ss(num,den);Af=flipud(Acc);A=fliplr(Af)B=flipud(Bcc)C=fliplr(Ccc)Computer response:ans =ans =aden =1 5 1 5 13Transfer function:8 s + 10----------------------------s^4 + 5 s^3 + s^2 + 5 s + 13A =0 1 0 00 0 1 00 0 0 1-13 -5 -1 -5
  • 82. 3-15 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.B =0001C =10 8 0 0ans =bden =1 9 13 8 0 0Transfer function:s^4 + 2 s^3 + 12 s^2 + 7 s + 6------------------------------s^5 + 9 s^4 + 13 s^3 + 8 s^2A =0 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 13 216 10 5s s s+ + +0 0 -8 -13 -9B =00001C =6 7 12 2 113.The transfer function can be represented as a block diagram as follows:1s3+6s2+9s+4R(s) X(s) Y(s)s2+3s+73 216 10 5s s s+ + +
  • 83. 3-16 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.Writing the differential equation for the first box:6 10 5 ( )x x x x r t••• •• •+ + + =Defining the state variables:123x xx xx x•••===Thus,1 22 33 1 2 35 10 6 ( ) 5 10 6 ( )x xx xx x x x r t x x x r t••• • ••=== − − − + = − − − +From the second box,1 2 33 8 8 3y x x x x x x•• •= + + = + +In vector-matrix form:[ ]0 1 0 00 0 1 0 ( )5 10 6 18 3 1r ty•⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦=x xx14.a. G(s)=C(sI-A)-1BA =0 1 00 0 1-3 -2 -5; B =0010; C = 1 0 0(sI - A)-1=1s3+ 5s2+ 2s +3s2+5s+2 s+5 1-3 s(s+5) s-3s -2s-3 s2Therefore, G(s) =10s3+5s2+2s+3.b. G(s)=C(sI-A)-1BA2 3 8−0 5 33− 5− 4−= ; B146= ; C 1 3 6, ,=
  • 84. 3-17 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.s I A− 1− 1s3 3 s2− 27 s− 157+s2 s− 5− 3 s 52+ 8 s− 49+9− s2 2 s 32−+ 3 s 6−3 s− 15+ 5 s− 1+ s2 7 s− 10+=Therefore, G s49 s2349 s− 452+s33 s2− 27 s− 157+= .c. G(s)=C(sI-A)-1BA =3 −5 21 −8 7−3 −6 2⎡⎣⎢⎢⎤⎦⎥⎥; B =5−32⎡⎣⎢⎢⎤⎦⎥⎥; C = 1 −4 3[ ](sI − A)−1=1s3+ 3s2+19s −133(s2+ 6s + 26) −(5s + 2) (2s −19)(s − 23) (s2− 5s +12) (7s −19)−(3s + 30) −(6s − 33) (s2+ 5s −19)⎡⎣⎢⎢⎤⎦⎥⎥Therefore, G(s) =23s2− 48s − 7s3+ 3s2+19s −133.15.Program:aA=[0 1 5 0;0 0 1 0;0 0 0 1;-7 -9 -2 -3];B=[0;5;8;2];C=[1 3 6 6];D=0;statespace=ss(A,B,C,D)[num,den]=ss2tf(A,B,C,D);G=tf(num,den)bA=[3 1 0 4 -2;-3 5 -5 2 -1;0 1 -1 2 8;-7 6 -3 -4 0;-6 0 4 -3 1];B=[2;7;8;5;4];C=[1 -2 -9 7 6];D=0;statespace=ss(A,B,C,D)[num,den]=ss2tf(A,B,C,D);G=tf(num,den)Computer response:ans =aa =x1 x2 x3 x4x1 0 1 5 0x2 0 0 1 0x3 0 0 0 1x4 -7 -9 -2 -3
  • 85. 3-18 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.b =u1x1 0x2 5x3 8x4 2c =x1 x2 x3 x4y1 1 3 6 6d =u1y1 0Continuous-time model.Transfer function:75 s^3 - 96 s^2 - 2331 s - 210------------------------------s^4 + 3 s^3 + 2 s^2 + 44 s + 7ans =ba =x1 x2 x3 x4 x5x1 3 1 0 4 -2x2 -3 5 -5 2 -1x3 0 1 -1 2 8x4 -7 6 -3 -4 0x5 -6 0 4 -3 1b =u1x1 2x2 7x3 8x4 5x5 4c =x1 x2 x3 x4 x5y1 1 -2 -9 7 6d =u1y1 0Continuous-time model.Transfer function:-25 s^4 - 292 s^3 + 1680 s^2 + 1.628e004 s + 3.188e004------------------------------------------------------s^5 - 4 s^4 - 32 s^3 + 148 s^2 - 1153 s - 448016.Program:syms saA=[0 1 3 00 0 1 0
  • 86. 3-19 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.0 0 0 1-7 -9 -2 -3];B=[0;5;8;2];C=[1 3 4 6];D=0;I=[1 0 0 00 1 0 00 0 1 00 0 0 1];T(s)T=C*((s*I-A)^-1)*B+D;T=simple(T);pretty(T)bA=[3 1 0 4 -2-3 5 -5 2 -10 1 -1 2 8-7 6 -3 -4 0-6 0 4 -3 1];B=[2;7;6;5;4];C=[1 -2 -9 7 6];D=0;I=[1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1];T(s)T=C*((s*I-A)^-1)*B+D;T=simple(T);pretty(T)Computer response:ans =aans =T(s)2 3-164 s - 1621 s - 260 + 59 s------------------------------4 3 2s + 3 s + 2 s + 30 s + 7ans =bans =T(s)2 3 414582 s + 1708 s - 408 s - 7 s + 27665------------------------------------------5 4 3 2s - 4 s - 32 s + 148 s - 1153 s - 448017.Let the input bedθzdt =ωz, x1=θx , x2=θ.x . Therefore,x.1 = x2
  • 87. 3-20 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.x.2 = -KxJxx1 -DxJxx2 + JωωzThe output is θx.In vector-matrix form, θx = x1 . Therefore, y = x1.x.=0 1-KxJx-DxJxx +0Jωω zy = 1 0 x18.The equivalent cascade transfer function is as shown below.KaK3s3+K2K3s2+K1K3s+K0K3s+KbKad (s) X(s) F (s)For the first box, x...+K2K3x..+K1K3x.+K0K3x =KaK3δ(t).Selecting the phase variables as the state variables: x1=x, x2=x., x3=x...Writing the state and output equations:x.1 = x2x.2 = x3x.3 = -K0K3x1-K1K3x2-K2K3x3+KaK3δ(t)y = φ(t) = x.+KbKax =KbKax1+x2In vector-matrix form,x.=0 1 00 0 1-K0K3-K1K3-K2K3x+00KaK3δ(t) ; y =KbKa1 0 x
  • 88. 3-21 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.19.Since Tm = Jeqdωmdt + Deqωm, and Tm = Kt ia,Jeqdωmdt + Deqωm = Kt ia (1)Or,dωmdt = -DeqJeqωm +KtJeqiaBut, ωm =N2N1ωL.Substituting in (1) and simplifying yields the first state equation,dωLdt = -DeqJeqωL +KtJeqN1N2iaThe second state equation is:dθLdt = ωLSinceea = Raia+Ladiadt +Kbωm = Raia+Ladiadt +KbN2N1ωL,the third state equation is found by solving fordiadt . Hence,diadt = -KbLaN2N1ωL -RaLaia+1LaeaThus the state variables are: x1 = ωL, x2 = θL , and x3 = ia.Finally, the output is y = θm =N2N1θL .In vector-matrix form,x.=-DeqJeq0KtJeqN1N21 0 0-KbLaN2N10 -RaLax+001Laea ; y = 0N2N10 xwhere,x =ωLθLia
  • 89. 3-22 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.20.Writing the differential equations,d2x1dt2 +dx1dt + 2x12 -dx2dt = 0d2x2dt2 +dx2dt -dx1dt = f(t)Defining the state variables to be x1, v1, x2, v2, where vs are velocity,x.1 = v1x.2 = v2v.1 = -v1-2x12+v2v.2 = v1-v2+f(t)Around x1 = 1, x1 = 1+δx1, and x.1 = δ x.1 . Also,x12= x12⎮x=1+dx1dt ⎮x=1δx1= 1+2x1 ⎮x=1δx1= 1+2δx1Therefore, the state and output equations are,δx.1 = v1x.2 = v2v.1 = -v1-2(1+2δx1)+v2v.2 = v1-v2+f(t)y = x2In vector-matrix form,δx1.x.2v.1v.2=0 0 1 00 0 0 1-4 0 -1 10 0 1 -1δx1x2v1v2+0 00 0-2 00 11f(t); y = 0 1 0 0δx1x2v1v2where f(t) = 2 + δf(t), since force in nonlinear spring is 2 N and must be balanced by 2 N force ondamper.21.Controller:The transfer function can be represented as a block diagram as follows:
  • 90. 3-23 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.Rc(s) Xc(s) Yc(s)K1s + bs+ aWriting the differential equation for the first box,K1s + band solving for cx•,1 ( )c c cx bx K r t•= − +From the second box,11( )( ) ( )c c c c c cc cy x ax bx K r t axa b x K r t•= + = − + += − +Wheels:The transfer function can be represented as a block diagram as follows:Rw(s) Xw(s)cs + cWriting the differential equation for the block of the form,cs + cand solving for wx•,( )w w wx cx cr t•= − +The output equation is,yw = xwVehicle:The transfer function can be represented as a block diagram as follows:Rv(s) Xv(s)1sWriting the differential equation for the block,
  • 91. 3-24 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.1sand solving for vx•,( )v vx r t•=The output equation isyv = xv22.A1.702− 50.72 263.380.22 1.418− 31.99−0 0 14−= ; B272.06−014=For G1(s), C 1 1 0 0, ,= , andG1(s) = C1(sI-A)-1BThus,G 1 C 11s 3 17.12 s 2 34.935 s 122.43−+ +s 2 15.418 s 19.852+ + 50.72 s 710.08+ 263.38 s 1249.1−0.22 s 3.08+ s 2 15.702 s 23.828+ + 31.99 s− 3.4966+0 0 s 2 3.12 s 8.745−+B=sOrG 1 s272.06 s 2− 507.3 s− 22888−s 3 17.12 s 2 34.935 s 122.43−+ += =272.06 s 2 1.8647 s 84.128+ +−s 14+ s 1.7834− s 4.9034+For G2(s), C2 = (0,1,0), andG2(s) = C2(sI-A)-1BThus,G 2 s C 21s 3 17.12 s 2 34.935 s 122.43−+ +s 2 15.418 s 19.852+ + 50.72 s 710.08+ 263.38 s 1249.1−0.22 s 3.08+ s 2 15.702 s 23.828+ + 31.99 s− 3.4966+0 0 s 2 3.12 s 8.745−+B=OrG 2 s507.71 s− 788.99−s 3 17.12 s 2 34.935 s 122.43−+ += =507.71 s 1.554+−s 14+ s 1.7834− s 4.9034+
  • 92. 3-25 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.23.Adding displacements to the figure,xezxsxrWriting the differential equations for noncontact,d2xrdt2 + 2dxrdt+ 2xr − xs −dxsdt= u(t)−dxrdt− xr +d2xsdt2 +dxsdt+ xs = 0Define the state variables as,x1 = xr ; x2 = xr•; x3 = xs ; x4 = xs•Writing the state equations, using the differential equations and the definition of the state variables,we get,x1•= xr•= x2x2•= xr••= −2x1 − 2x2 + x3 + x4 + u(t)x3•= xs•= x4x4•= xs••= x1 + x2 − x3 − x4Assuming the output to be xs, the output equation is,y = x3In vector-matrix form,
  • 93. 3-26 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.x•=0 1 0 0−2 −2 1 10 0 0 11 1 −1 −1⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥x +0100⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥u(t)y = 0 0 1 0[ ]xWriting the differential equations for contact,d2xrdt2+ 2dxrdt+ 2xr − xs −dxsdt= u(t)−dxrdt− xr +d2xsdt2+dxsdt+ xs − z − xe = 0−xs +dzdt+ z −dxedt= 0−xs −dzdt+d2xedt2+ 2dxedt+ 2xe = 0Defining the state variables,x1 = xr ; x2 = xr•; x3 = xs ; x4 = xs•; x5 = z; x6 = z•; x7 = xe; x8 = xe•Using the differential equations and the definitions of the state variables, we write the state equations.x1•= x2x2•= −x1 − 2x2 + x3 + x4 +u(t)x3•= x4x4•= x1 + x2 − x3 − x4 + x5 + x7x5•= x6Differentiating the third differential equation and solving for d2z/dt2we obtain,x6•=d2zdt2 =dxsdt−dzdt+d2xedt2But, from the fourth differential equation,d2xedt2 = xs +dzdt− 2dxedt− 2xe = x3 + x6 − 2x8 − 2x7Substituting this expression back into x6•along with the other definitions and then simplifying yields,x6•= x4 + x3 − 2x8 − 2x7Continuing,
  • 94. 3-27 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.x7•= x8x8•= x3 + x6 − 2x7 − 2x8Assuming the output is xs,y = xsHence, the solution in vector-matrix form isx•=0 1 0 0 0 0 0 0−1 −2 1 1 0 0 0 00 0 0 1 0 0 0 01 1 −1 −1 1 0 1 00 0 0 0 0 1 0 00 0 1 1 0 0 −2 −20 0 0 0 0 0 0 10 0 1 0 0 1 −2 −2⎡⎣⎢⎢⎢⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎥⎥⎥x +01000000⎡⎣⎢⎢⎢⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎥⎥⎥u(t)y = 0 0 1 0 0 0 0 0[ ]x24.a. We begin by calculating⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛+−+−−−+=−02.00227.00394.0268.002.0209.0435.0sssAsIand0227.0394.0268.002.002.0227.00268.0209.002.000394.0)435.0()det(−+−−+−−++++=−sssssAsI)394.0)(227.0(02.0)02.0)(268.0(209.0)02.0)(394.0)(435.0( +−+−++++= sssss00179.000454.000112.0056.00034.0188.0849.0 23−−−−+++= sssss)004.0)(19.0)(66.0(00049.01278.0849.0 23+++=+++= ssssss
  • 95. 3-28 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=−332313322212312111)(cccccccccAdj AsIwhere)02.0)(394.0(02.000394.011 ++=++= ssssc)02.0(268.002.0227.00268.012 +−=+−−= ssc)394.0(227.00227.0394.0268.013 +−=−+−= ssc)02.0(209.002.0002.0209.021 +−=+−−= ssc00416.0455.000454.0)02.0)(435.0(02.0227.002.0435.0 222 ++=−++=+−−+= ssssssc047443.00227.0209.0435.023 −=−−+=sc)394.0(02.00394.002.0209.031 +=+−−= ssc00536.00268.002.0435.032 −=−−+=sc1154.0829.0)209.0(268.0)394.0)(435.0(394.0268.0209.0435.0 233 ++=−++=+−−+= ssssssc)66.0)(19.0)(004.0(1154.0829.00474.0)394.0(227.00054.00042.0455.0)02.0(268.0)394.0(02.0)02.0(209.0)02.0)(394.0()det()()(22+++⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++−+−−+++−++−++=−−=− −sssssssssssssAdjAsIAsIAsI 1[ ])66.0)(19.0)(004.0(33.3333)394.0(02.0)02.0(209.0)02.0)(394.0()(+++++−++=− −sssssss1AsIC)66.0)(19.0)(004.0(33.3333)394.0)(02.0()()()(+++++=−= −ssssssUsYBAsIC 1
  • 96. 3-29 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.b.>> A=[-0.435 0.209 0.02; 0.268 -0.394 0; 0.227 0 -0.02]A =-0.4350 0.2090 0.02000.2680 -0.3940 00.2270 0 -0.0200>> B = [1;0;0]B =100>> C = [0.0003 0 0]C =1.0e-003 *0.3000 0 0>> [n,d]=ss2tf(A,B,C,0)
  • 97. 3-30 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.n =1.0e-003 *0 0.3000 0.1242 0.0024d =1.0000 0.8490 0.1274 0.0005>> roots(n)ans =-0.3940-0.0200>> roots(d)ans =-0.6560-0.1889-0.0042
  • 98. 3-31 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.25. By direct observation0432104442333224232221201211100200432100000100000000000dxxxxxaaaaaaaaaaaaaaxxxxx⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡&&&&&26.a.>> A=[-0.038 0.896 0 0.0015; 0.0017 -0.092 0 -0.0056; 1 0 0 -3.086; 0 1 0 0]A =-0.0380 0.8960 0 0.00150.0017 -0.0920 0 -0.00561.0000 0 0 -3.08600 1.0000 0 0>> B = [-0.0075 -0.023; 0.0017 -0.0022; 0 0; 0 0]B =-0.0075 -0.02300.0017 -0.00220 00 0>> C = [0 0 1 0; 0 0 0 1]
  • 99. 3-32 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.C =0 0 1 00 0 0 1>> [num,den] = ss2tf(A,B,C,zeros(2),1)num =0 0.0000 -0.0075 -0.0044 -0.00020 0 0.0017 0.0001 0den =1.0000 0.1300 0.0076 0.0002 0>> [num,den] = ss2tf(A,B,C,zeros(2),2)num =0 -0.0000 -0.0230 0.0027 0.00020 -0.0000 -0.0022 -0.0001 0den =1.0000 0.1300 0.0076 0.0002 0
  • 100. 3-33 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.b.From the MATLAB results)0002.00076.013.0()0002.00044.00075.0()( 232+++++−=ssssssszBδ0002.00076.013.00001.00017.0)( 23++++=sssssBδθ)0002.00076.013.0(0002.00027.0023.0)( 232+++++−=ssssssszSδ0002.00076.013.0)0001.00022.0()( 23++++−=sssssSδθ27.The transfer function is divided into two parts:bdsadbsdas +++++ )()(123cs +So we havebdsadbsdassRsW+++++=)()(1)()(23and cssWsY+=)()(In time domainrbdwwadbwdaw =+++++ &&&&&& )()( and ycww =+&
  • 101. 3-34 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.Define the state variables as23121xwxxwxwx&&&&&=====So we can write3213 )()( xdaxadbbdxrxw +−+−−== &&&& and 21 xcxy +=In matrix form these equations are:rxxxdaadbbdxxx⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+−+−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡100)()(100010321321&&&[ ]⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=32101xxxcy28.a.)5(213)5(7)( 1+=+=−= −−sssG BA)C(sI 1b.[ ] [ ] ⎥⎦⎤⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++=⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡++=−=−−131100510713100507)(1sssssG BA)C(sI 152113057+=⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡+=ss
  • 102. 3-35 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.c) [ ] [ ] ⎥⎦⎤⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++=⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡++=−=−−031100513703100537)(1sssssG BA)C(sI 1[ ]52105307+=⎥⎥⎦⎤⎢⎢⎣⎡+=ss29.a.  1 2 3 4( ) ( ) ( ) ( )SOO A O O SO O IDOdmk m t k k m t k m tdt= − + +  3 4( ) ( )IDOO SO O IDOdmk m t k m tdt= −  1 2 3( ) ( ) ( )VL A L L Vdmk m t k k m tdt= − +  3 4( ) ( )SL V L Sdmk m t k m tdt= −              b.  01 1 02 2 402 02 03 0402 041 2 32 4( ) 0 1( ) 0 0 0( )0 0 0 00 0 ( ) 0 00 0 0 0AAL L LSOSOEIDOIDOVL L LVSL LSmmk k k k km mk k k ku tmk kmmk k kmmk km•••••⎡ ⎤⎢ ⎥ − + ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ − + ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= +−⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ − + ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦ 
  • 103. 3-36 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.[ ]1 0 0 0 0ASOIDOVSmmmmm⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦y30.Writing the equations of motion,Mfd2yfdt2+( fvf + fvh )dyfdt+ Kh yf − fvhdyhdt− Kh yh = fup(t)− fvhdyfdt− Khyf + Mhd 2yhdt2 + fvhdyhdt+(Kh + Ks )yh − Ks ycat = 0−Ks yh + (Ks + Kave )ycat = 0The last equation says thatycat =Ks(Ks + Kave )yhDefining state variables for the first two equations of motion,x1 = yh ; x2 = yh•; x3 = yf ; x4 = yf•Solving for the highest derivative terms in the first two equations of motion yields,d2yfdt2 = −( fvf + fvh )M fdyfdt−KhM fyf +fvhMfdyhdt+KhMfyh +1Mffup(t)d2yhdt2=fvhMhdyfdt+KhMhyf −fvhMhdyhdt−(Kh + Ks )Mhyh +KsMhycatWriting the state equations,x1•= x2x2•=fvhMhx4 +KhMhx3 −fvhMhx2 −(Kh + Ks )Mhx1 +KsMhKs(Ks + Kave )x1x3•= x4x4•= −( fvf + fvh )Mfx4 −KhMfx3 +fvhM fx2 +KhMfx1 +1Mffup (t)The output is yh - ycat. Therefore,
  • 104. 3-37 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.y = yh − ycat = yh −Ks(Ks + Kave )yh =Kave(Ks + Kave )x1Simplifying, rearranging, and putting the state equations in vector-matrix form yields,x•=0 1 0 01MhKs2(Ks + Kave )− (Kh + Ks )⎛⎝⎜ ⎞⎠⎟ −fvhMhKhMhfvhMh0 0 0 1KhM ffvhMf−KhMf−( fvf + fvh )Mf⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥x +0001Mf⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥fup (t)y =Kave(Ks + Kave )0 0 0⎡⎣⎢⎤⎦⎥xSubstituting numerical values,x•=0 1 0 0−9353 −14.29 769.2 14.290 0 0 1406 7.558 −406 −9.302⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥x +0000.0581⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥fup(t)y = 0.9491 0 0 0[ ]x31.a.TvudTsf β)1( 11 −−−=*12 )1( TTvuf μβ −−=cvkTuf −−= *23 )1(0100101)1(|)1( vudvudTfββ −−−=−−−=∂∂
  • 105. 3-38 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.00*1=∂∂Tf0100101)1(|)1( vuvuvfββ −=−=∂∂0100102)1(|)1( vuvuTfββ −=−=∂∂μ−=∂∂0*2Tf0100102)1(|)1( TuTuvfββ −=−=∂∂003=∂∂TfkuTf)1( 200*3−=∂∂cvf−=∂∂0300011vTufβ=∂∂0021=∂∂uf00012vTufβ−=∂∂0022=∂∂uf0013=∂∂uf*0023kTuf−=∂∂Then just by direct substitution.
  • 106. 3-39 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.b.Substituting values one gets:⎥⎦⎤⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−+−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡21*00000*0000*00000)(uukTvTvTvTTckTvTvdvTTβββμβββ&&&[ ]⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=vTTy *10032.a. The following basic equations characterize the relationships between the state, input, and outputvariables for the HEV common forward path of the figure:)()( tuKtu cAa ⋅=)()()()()( tktuKtetutIRIL bcAbaaaaA ω⋅−=−=⋅+⋅ & (1))()()( tTtTtTJ cftot −−=⋅ω& , where tot m veh wJ J J J= + + ,)()( tIktT at ⋅= , )()( tktT ff ω⋅=b. Given that the state variables are the motor armature current, Ia(t), and angular speed, ω (t), we re-write the above equations as:)()()( tuLKtLktILRI caAabaaaa +⋅−⋅−= ω& (2))(1)()( tTJtJktIJkctottotfatott−−= ωω& (3)In matrix form, the resulting state-space equations are:
  • 107. 3-40 Chapter 3: Modeling in the Time DomainCopyright © 2011 by John Wiley & Sons, Inc.⎥⎦⎤⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−+⎥⎦⎤⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−=⎥⎦⎤⎢⎣⎡cctotaAatotftottabaaaTuJLKIJkJkLkLRI100ωω&& (4)⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡ωωaa II1001(5)
  • 108. Copyright © 2011 by John Wiley & Sons, Inc.F O U RTime ResponseSOLUTIONS TO CASE STUDIES CHALLENGESAntenna Control: Open-Loop ResponseThe forward transfer function for angular velocity is,G(s) =ω0(s)VP(s) =24(s+150)(s+1.32)a. ω0(t) = A + Be-150t + Ce-1.32tb. G(s) =24s2+151.32s+198. Therefore, 2ζωn =151.32, ωn = 14.07, and ζ = 5.38.c. ω0(s) =24s(s2+151.32s+198)=Therefore, ω0(t) = 0.12121 + .0010761 e-150t - 0.12229e-1.32t.d. Using G(s),ω0••+151.32ω0•+198ω0 = 24vp (t)Defining,x1 = ω0x2 = ω0•Thus, the state equations are,x1•= x2x2•= −198x1 −151.32x2 + 24vp(t)y = x1In vector-matrix form,x•=0 1−198 −151.32⎡⎣⎢⎤⎦⎥x +024⎡⎣⎢⎤⎦⎥vp (t); y = 1 0[ ]x
  • 109. 4-2 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.e.Program:Case Study 1 Challenge (e)num=24;den=poly([-150 -1.32]);G=tf(num,den)step(G)Computer response:ans =Case Study 1 Challenge (e)Transfer function:24-------------------s^2 + 151.3 s + 198Ship at Sea: Open-Loop Responsea. Assuming a second-order approximation: ωn2 = 2.25, 2ζωn = 0.5. Therefore ζ = 0.167, ωn = 1.5.Ts =4ζωn= 16; TP =πωn 1-ζ2= 2.12 ;%OS = e-ζπ / 1 - ζ2x 100 = 58.8%; ωnTr = 1.169 therefore, Tr = 0.77.b. θ s 2.25s s 2 0.5 s 2.25+ += = 1ss 0.5+s 2 0.5 s 2.25+ +−= 1ss 0.25+ 0.252.18752.1875+s 0.25+ 2 2.1875+−
  • 110. Answers to Review Questions 4-3Copyright © 2011 by John Wiley & Sons, Inc.= 1ss 0.25+ 0.16903 1.479+s 0.25+ 2 2.1875+−Taking the inverse Laplace transform,θ(t) = 1 - e-0.25t ( cos1.479t +0.16903 sin1.479t)c.Program:Case Study 2 Challenge (C)(a)numg=2.25;deng=[1 0.5 2.25];G=tf(numg,deng)omegan=sqrt(deng(3))zeta=deng(2)/(2*omegan)Ts=4/(zeta*omegan)Tp=pi/(omegan*sqrt(1-zeta^2))pos=exp(-zeta*pi/sqrt(1-zeta^2))*100t=0:.1:2;[y,t]=step(G,t);Tlow=interp1(y,t,.1);Thi=interp1(y,t,.9);Tr=Thi-Tlow(b)numc=2.25*[1 2];denc=conv(poly([0 -3.57]),[1 2 2.25]);[K,p,k]=residue(numc,denc)(c)[y,t]=step(G);plot(t,y)title(Roll Angle Response)xlabel(Time(seconds))ylabel(Roll Angle(radians))Computer response:ans =Case Study 2 Challenge (C)ans =(a)Transfer function:2.25------------------s^2 + 0.5 s + 2.25omegan =1.5000zeta =0.1667Ts =16
  • 111. 4-4 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.Tp =2.1241pos =58.8001Tr =0.7801ans =(b)K =0.1260-0.3431 + 0.1058i-0.3431 - 0.1058i0.5602p =-3.5700-1.0000 + 1.1180i-1.0000 - 1.1180i0k =[]ans =(c)
  • 112. Answers to Review Questions 4-5Copyright © 2011 by John Wiley & Sons, Inc.ANSWERS TO REVIEW QUESTIONS1.Time constant2. The time for the step response to reach 63% of its final value3. The input pole4. The system poles5. The radian frequency of a sinusoidal response6. The time constant of an exponential response7. Natural frequency is the frequency of the system with all damping removed; the damped frequency ofoscillation is the frequency of oscillation with damping in the system.8. Their damped frequency of oscillation will be the same.9. They will all exist under the same exponential decay envelop.10. They will all have the same percent overshoot and the same shape although differently scaled in time.11. ζ, ωn, TP, %OS, Ts12. Only two since a second-order system is completely defined by two component parameters13. (1) Complex, (2) Real, (3) Multiple real14. Poles real part is large compared to the dominant poles, (2) Pole is near a zero15. If the residue at that pole is much smaller than the residues at other poles16. No; one must then use the output equation17. The Laplace transform of the state transition matrix is (sI -A)-118. Computer simulation19. Pole-zero concepts give one an intuitive feel for the problem.20. State equations, output equations, and initial value for the state-vector21. Det(sI-A) = 0SOLUTIONS TO PROBLEMS1.a. Overdamped Case:C(s) =9s(s2 + 9s + 9)Expanding into partial fractions,9 1 0.171 1.171(s) -s(s 7.854)(s 1.146) s (s 7.854) (s 1.146)C = = ++ + + +Taking the inverse Laplace transform,c(t) = 1 + 0.171 e-7.854t - 1.171 e-1.146t
  • 113. 4-6 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.b. Underdamped Case:K2 and K3 can be found by clearing fractions with K1 replaced by its value. Thus,9 = (s2 + 3s + 9) + (K2s + K3)sor9 = s2 + 3s +9 + K2s2 + K3sHence K2 = -1 and K3 = -3. Thus,c(t) = 1 -23e-3t/2 cos(274 t - φ)= 1 - 1.155 e -1.5t cos (2.598t - φ)whereφ = arctan (327) = 30o
  • 114. Solutions to Problems 4-7Copyright © 2011 by John Wiley & Sons, Inc.c. Oscillatory Case:The evaluation of the constants in the numerator are found the same way as they were for theunderdamped case. The results are K2 = -1 and K3 = 0. Hence,Therefore,c(t) = 1 - cos 3td. Critically DampedThe constants are then evaluated asNow, the transform of the response isc(t) = 1 - 3t e-3t - e-3t2.a. C(s) =5s(s+5) =1s -1s+5 . Therefore, c(t) = 1 - e-5t.Also, T =15 , Tr =2.2a =2.25 = 0.44, Ts =4a =45 = 0.8.b. C(s) =20s(s+20) =1s -1s+20 . Therefore, c(t) = 1 - e-20t. Also, T =120 ,Tr =2.2a =2.220 = 0.11, Ts =4a =420 = 0.2.
  • 115. 4-8 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.3.Program:(a)num=5;den=[1 5];Ga=tf(num,den)subplot(1,2,1)step(Ga)title((a))(b)num=20;den=[1 20];Gb=tf(num,den)subplot(1,2,2)step(Gb)title((b))Computer response:ans =(a)Transfer function:5-----s + 5ans =(b)Transfer function:20------s + 20
  • 116. Solutions to Problems 4-9Copyright © 2011 by John Wiley & Sons, Inc.4.Using voltage division,VC(s)Vi(s) =1/ 0.7031 0.703RCsSRC=++. Since Vi(s) =5s5 0.703 5 5( )0.703 0.703cV ss s s s⎛ ⎞= = −⎜ ⎟+ +⎝ ⎠.Therefore0.703( ) 5 5 tcv t e−= − . Also,s1 2.2 41.422; = 3.129; T 5.690.703 0.703 0.703rT T= = = = = .5.Program:clfnum=0.703;den=[1 0.703];G=tf(num,den)step(5*G)Computer response:Transfer function:0.703---------s + 0.703
  • 117. 4-10 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.6.Writing the equation of motion,2( 6 ) ( ) ( )Ms s X s F s+ =Thus, the transfer function is,2( ) 1( ) 6X sF s Ms s=+Differentiating to yield the transfer function in terms of velocity,( ) 1 1/6( ) 6sX s MF s Ms sM= =+ +Thus, the settling time, Ts, and the rise time, Tr, are given by4 2 2.2 1.10.667 ; 0.3676/ 3 6/ 3s rT M M T M MM M= = = = = =Tc = 1.4
  • 118. Solutions to Problems 4-11Copyright © 2011 by John Wiley & Sons, Inc.7.Program:ClfM=1num=1/M;den=[1 6/M];G=tf(num,den)step(G)pauseM=2num=1/M;den=[1 6/M];G=tf(num,den)step(G)Computer response:M =1Transfer function:1-----s + 6M =2Transfer function:0.5-----s + 3
  • 119. 4-12 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.From plot, time constant =.0.16 s.Tc
  • 120. Solutions to Problems 4-13Copyright © 2011 by John Wiley & Sons, Inc.From plot, time constant = 0.33 s.8.a. Pole: -2; c(t) = A + Be-2t ; first-order response.b. Poles: -3, -6; c(t) = A + Be-3t + Ce-6t; overdamped response.c. Poles: -10, -20; Zero: -7; c(t) = A + Be-10t + Ce-20t; overdamped response.d. Poles: (-3+j3 15 ), (-3-j3 15 ) ; c(t) = A + Be-3t cos (3 15 t + φ); underdamped.e. Poles: j3, -j3; Zero: -2; c(t) = A + B cos (3t + φ); undamped.f. Poles: -10, -10; Zero: -5; c(t) = A + Be-10t + Cte-10t; critically damped.9.Program:p=roots([1 6 4 7 2])Computer response:p =Tc
  • 121. 4-14 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.-5.4917-0.0955 + 1.0671i-0.0955 - 1.0671i-0.317310.G(s) = C (sI-A)-1 B[ ]8 4 1 43 2 0 ; 3 ; 2 8 35 7 9 4− −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − = − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦A B C1 23 22( 2)( 9) (4 29) ( 2)1( ) (3 27) ( 77) 391 675 31 7 76 ( 10 4)s s s ss s s ss s ss s s s−− + − + −⎡ ⎤⎢ ⎥− = − + + − −⎢ ⎥− − +⎢ ⎥− − − +⎣ ⎦I ATherefore, G(s ) =23 244 291 181491 67s ss s s− + +− − +.Factoring the denominator, or using det(sI-A), we find the poles to be 9.683, 0.7347, -9.4179.11.Program:A=[8 -4 1;-3 2 0;5 7 -9]B=[-4;-3;4]C=[2 8 -3]D=0[numg,deng]=ss2tf(A,B,C,D,1);G=tf(numg,deng)poles=roots(deng)Computer response:A =8 -4 1-3 2 05 7 -9B =-4-34
  • 122. Solutions to Problems 4-15Copyright © 2011 by John Wiley & Sons, Inc.C =2 8 -3D =0Transfer function:-44 s^2 + 291 s + 1814----------------------s^3 - s^2 - 91 s + 67poles =-9.41799.68320.734712.Writing the node equation at the capacitor, VC(s) (1R2+1Ls + Cs) +VC(s) - V(s)R1= 0.Hence,VC(s)V(s) =1R11R1+1R2+1Ls + Cs=10ss2+20s+500. The step response is10s2+20s+500.The polesare at-10 ± j20. Therefore, vC(t) = Ae-10t cos (20t + φ).13.Program:num=[10 0];den=[1 20 500];G=tf(num,den)step(G)Computer response:Transfer function:10 s----------------s^2 + 20 s + 500
  • 123. 4-16 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.14.The equation of motion is: (Ms2+fvs+Ks)X(s) = F(s). Hence,X(s)F(s) =1Ms2+fvs+Ks=1s2+s+5.The step response is now evaluated: X(s) =1s(s2+s+5)=1/5s -15 s +15(s+12)2+194=15(s+12) +15 19192(s+12)2 +194.Taking the inverse Laplace transform, x(t) =15 -15 e-0.5t ( cos192 t +119sin192 t )=15 ⎣⎡⎦⎤1 - 2519 e-0.5t cos (192 t - 12.92o) .15.C(s) =ωn2s(s2+2ζωns+ωn2)=1s -s + 2ζωns2+2ζωns+ωn2 =1s -s + 2ζωn(s+ζωn)2 + ωn2 - ζ2ωn2=1s -(s + ζωn) + ζωn(s+ζωn)2 + (ωn 1 - ζ2)2=1s -(s+ζωn) +ζωnωn 1 - ζ2ωn 1 - ζ2(s+ζωn)2 + (ωn 1 - ζ2)2Hence, c(t) = 1 - e-ζω ntcos ωn 1 - ζ 2t +ζ1 - ζ 2sin ωn 1 - ζ 2t⎛⎝⎜⎞⎠⎟
  • 124. Solutions to Problems 4-17Copyright © 2011 by John Wiley & Sons, Inc.= 1 - e-ζωnt 1 +ζ21-ζ2 cos (ωn 1 - ζ2 t - φ) = 1 - e-ζωnt 11-ζ2cos (ωn 1 - ζ2 t - φ),where φ = tan-1 ζ1 - ζ216.%OS = e-ζπ / 1 - ζ2x 100. Dividing by 100 and taking the natural log of both sides,ln (%OS100 ) = -ζπ1 - ζ2. Squaring both sides and solving for ζ2, ζ2 =ln2 (%OS100 )π2 + ln2 (%OS100 ). Taking thenegative square root, ζ =- ln (%OS100 )π2 + ln2 (%OS100 ).17.a.b.c.d.
  • 125. 4-18 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.e.f.18.a. N/Ab. s2+9s+18, ωn2 = 18, 2ζωn = 9, Therefore ζ = 1.06, ωn = 4.24, overdamped.c. s2+30s+200, ωn2 = 200, 2ζωn = 30, Therefore ζ = 1.06, ωn = 14.14, overdamped.d. s2+6s+144, ωn2 = 144, 2ζωn = 6, Therefore ζ = 0.25, ωn = 12, underdamped.e. s2+9, ωn2 = 9, 2ζωn = 0, Therefore ζ = 0, ωn = 3, undamped.f. s2+20s+100, ωn2 = 100, 2ζωn = 20, Therefore ζ = 1, ωn = 10, critically damped.19.X(s) =1002s(s2 +100s+1002)=1s -s+100(s+50)2+7500=1s -(s+50) + 50(s+50)2+7500=1s -(s+50) +5075007500(s+50)2+7500Therefore, x(t) = 1 - e-50t (cos 7500 t +507500sin 7500 t)= 1 -23e-50t cos (50 3 t - tan-1 13)20.a. ωn2 = 16 r/s, 2ζωn = 3. Therefore ζ = 0.375, ωn = 4. Ts =4ζωn= 2.667 s; TP =πωn 1-ζ2=0.8472 s; %OS = e-ζπ / 1 - ζ2x 100 = 28.06 %; ωnTr = (1.76ζ3- 0.417ζ2+ 1.039ζ + 1) = 1.4238;therefore, Tr = 0.356 s.
  • 126. Solutions to Problems 4-19Copyright © 2011 by John Wiley & Sons, Inc.b. ωn2 = 0.04 r/s, 2ζωn = 0.02. Therefore ζ = 0.05, ωn = 0.2. Ts =4ζωn= 400 s; TP =πωn 1-ζ2=15.73 s; %OS = e-ζπ / 1 - ζ2x 100 = 85.45 %; ωnTr = (1.76ζ3- 0.417ζ2+ 1.039ζ + 1); therefore,Tr = 5.26 s.c. ωn2 = 1.05 x 107r/s, 2ζωn = 1.6 x 103. Therefore ζ = 0.247, ωn = 3240. Ts =4ζωn= 0.005 s; TP =πωn 1-ζ2= 0.001 s; %OS = e-ζπ / 1 - ζ2x 100 = 44.92 %; ωnTr = (1.76ζ3- 0.417ζ2+ 1.039ζ +1); therefore, Tr = 3.88x10-4 s.21.Program:(a)clfnuma=16;dena=[1 3 16];Ta=tf(numa,dena)omegana=sqrt(dena(3))zetaa=dena(2)/(2*omegana)Tsa=4/(zetaa*omegana)Tpa=pi/(omegana*sqrt(1-zetaa^2))Tra=(1.76*zetaa^3 - 0.417*zetaa^2 + 1.039*zetaa + 1)/omeganapercenta=exp(-zetaa*pi/sqrt(1-zetaa^2))*100subplot(221)step(Ta)title((a))(b)numb=0.04;denb=[1 0.02 0.04];Tb=tf(numb,denb)omeganb=sqrt(denb(3))zetab=denb(2)/(2*omeganb)Tsb=4/(zetab*omeganb)Tpb=pi/(omeganb*sqrt(1-zetab^2))Trb=(1.76*zetab^3 - 0.417*zetab^2 + 1.039*zetab + 1)/omeganbpercentb=exp(-zetab*pi/sqrt(1-zetab^2))*100subplot(222)step(Tb)title((b))(c)numc=1.05E7;denc=[1 1.6E3 1.05E7];Tc=tf(numc,denc)omeganc=sqrt(denc(3))zetac=denc(2)/(2*omeganc)Tsc=4/(zetac*omeganc)Tpc=pi/(omeganc*sqrt(1-zetac^2))Trc=(1.76*zetac^3 - 0.417*zetac^2 + 1.039*zetac + 1)/omegancpercentc=exp(-zetac*pi/sqrt(1-zetac^2))*100subplot(223)step(Tc)title((c))
  • 127. 4-20 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.Computer response:ans =(a)Transfer function:16--------------s^2 + 3 s + 16omegana =4zetaa =0.3750Tsa =2.6667Tpa =0.8472Tra =0.3559percenta =28.0597ans =(b)Transfer function:0.04-------------------s^2 + 0.02 s + 0.04omeganb =0.2000zetab =0.0500Tsb =400
  • 128. Solutions to Problems 4-21Copyright © 2011 by John Wiley & Sons, Inc.Tpb =15.7276Trb =5.2556percentb =85.4468ans =(c)Transfer function:1.05e007-----------------------s^2 + 1600 s + 1.05e007omeganc =3.2404e+003zetac =0.2469Tsc =0.0050Tpc =0.0010Trc =3.8810e-004percentc =44.9154
  • 129. 4-22 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.22.Program:T1=tf(16,[1 3 16])T2=tf(0.04,[1 0.02 0.04])T3=tf(1.05e7,[1 1.6e3 1.05e7])ltiviewComputer response:Transfer function:16--------------s^2 + 3 s + 16Transfer function:0.04-------------------s^2 + 0.02 s + 0.04Transfer function:1.05e007-----------------------s^2 + 1600 s + 1.05e007
  • 130. Solutions to Problems 4-23Copyright © 2011 by John Wiley & Sons, Inc.
  • 131. 4-24 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.23.a. ζ =- ln (%OS100 )π2 + ln2 (%OS100 )= 0.56, ωn =4ζTs= 11.92. Therefore, poles = -ζωn ± jωn 1-ζ2= -6.67 ± j9.88.b. ζ =- ln (%OS100 )π2 + ln2 (%OS100 )= 0.591, ωn =πTP 1-ζ2= 0.779.Therefore, poles = -ζωn ± jωn 1-ζ2 = -0.4605 ± j0.6283.c. ζωn =4Ts= 0.571, ωn 1-ζ2 =πTp= 1.047. Therefore, poles = -0.571 ± j1.047.24.Re =4Ts= 4; ζ =-ln(12.3/100)π 2+ ln2(12.3/100)= 0.5549Re =ζωn = 0.5549ωn = 4; ∴ωn = 7.21Im = ωn 1 −ζ 2= 6∴G(s) =ωn2s2+ 2ζωns + ωn2 =51.96s2+ 8s + 51.96
  • 132. Solutions to Problems 4-25Copyright © 2011 by John Wiley & Sons, Inc.25.a. Writing the equation of motion yields,2(5 5 28) ( ) ( )s s X s F s+ + =Solving for the transfer function,2( ) 1/528( )5X sF s s s=+ +b. ωn2 = 28/5 r/s, 2ζωn = 1. Therefore ζ = 0.211, ωn = 2.37. Ts =4ζωn= 8.01 s; TP =πωn 1-ζ2=1.36 s; %OS = e-ζπ / 1 - ζ2x 100 = 50.7 %; ωnTr = (1.76ζ3- 0.417ζ2+ 1.039ζ + 1); therefore, Tr= 0.514 s.26.Writing the loop equations,21 21 2(1.07 1.53 ) ( ) 1.53 ( ) ( )1.53 ( ) (1.53 1.92) ( ) 0s s s s T ss s s sθ θθ θ+ − =− + + =Solving for θ2(s),22 22(1.07 1.53 ) ( )1.53 0 0.935 ( )( )1.25 1.79(1.07 1.53 ) 1.531.53 (1.53 1.92)s s T ss T sss ss s ss sθ+−= =+ ++ −− +Forming the transfer function,22( ) 0.935( ) 1.25 1.79sT s s sθ=+ +Thus ωn = 1.34, 2ζωn = 1.25. Thus, ζ = 0.467. From Eq. (4.38), %OS = 19.0%. From Eq. (4.42), Ts= 6.4 seconds. From Eq. (4.34), Tp = 2.66 seconds.27.a.24.542s(s2+ 4s + 24.542)=1s-s + 4(s + 2)2+ 20.542=1s-(s + 2) +24.5324.532(s + 2)2+ 20.542.Thus c(t) = 1 - e-2t (cos4.532t+0.441 sin 4.532t) = 1-1.09e-2t cos(4.532t -23.80).b.
  • 133. 4-26 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.Therefore, c(t) = 1 - 0.29e-10t - e-2t(0.71 cos 4.532t + 0.954 sin 4.532t)= 1 - 0.29e-10t - 1.189 cos(4.532t - 53.34o).c.Therefore, c(t) = 1 - 1.14e-3t + e-2t (0.14 cos 4.532t - 0.69 sin 4.532t)= 1 - 1.14e-3t + 0.704 cos(4.532t +78.53o).28.Since the third pole is more than five times the real part of the dominant pole, s2+0.842s+2.829determines the transient response. Since 2ζωn = 0.842, and ωn = 2.829 = ωn = 1.682, ζ = 0.25,2/ 1%OS x100 44.4%e ζπ ζ− −= = , Ts =4ζωn= 9.50 sec, Tp =πωn 1-ζ2= 1.93 sec; ωnTr =(1.76ζ3- 0.417ζ2+ 1.039ζ + 1) = 1.26, therefore, Tr = 0.75.
  • 134. Solutions to Problems 4-27Copyright © 2011 by John Wiley & Sons, Inc.29.a. Measuring the time constant from the graph, T = 0.0244 seconds.01230 0.05 0.1 0.15 0.2 0.25Time(seconds)T = 0.0244 secondsResponseEstimating a first-order system, G(s) =Ks+a . But, a = 1/T = 40.984, andKa = 2. Hence, K = 81.967.Thus,G(s) =81.967s+40.984b. Measuring the percent overshoot and settling time from the graph: %OS = (13.82-11.03)/11.03 =25.3%,05101520250 1 2 3 4 5ResponseTs = 2.62 secondscmax = 13.82cfinal = 11.03Time(seconds)
  • 135. 4-28 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.and Ts = 2.62 seconds. Estimating a second-order system, we use Eq. (4.39) to find ζ = 0.4 , and Eq.(4.42) to find ωn = 3.82. Thus, G(s) =Ks2+2ζωns +ωn2 . Since Cfinal = 11.03,Kωn2 = 11.03. Hence,K = 160.95. Substituting all values,G(s) =160.95s2+3.056s+14.59c. From the graph, %OS = 40%. Using Eq. (4.39), ζ = 0.28. Also from the graph,Tp =πωn 1 −ζ 2= 4. Substituting ζ = 0.28, we find ωn = 0.818.Thus,G(s) =Ks2+2ζωns +ωn2 =0.669s2+ 0.458s + 0.669.30.a.Since the amplitude of the sinusoids are of the same order of magnitude as the residue of the pole at -2, pole-zero cancellation cannot be assumed.b.Since the amplitude of the sinusoids are of the same order of magnitude as the residue of the pole at -2, pole-zero cancellation cannot be assumed.c.
  • 136. Solutions to Problems 4-29Copyright © 2011 by John Wiley & Sons, Inc.Since the amplitude of the sinusoids are of two orders of magnitude larger than the residue of the poleat -2, pole-zero cancellation can be assumed. Since 2ζωn = 1, and ωn = 5 = 2.236, ζ = 0.224,%OS = e−ζπ / 1−ζ2x100 = 48.64%, Ts =4ζωn= 8 sec, Tp =πωn 1-ζ2= 1.44 sec; ωnTr = 1.23,therefore, Tr = 0.55.d.Since the amplitude of the sinusoids are of two orders of magnitude larger than the residue of the poleat -2, pole-zero cancellation can be assumed. Since 2ζωn = 5, and ωn = 20 = 4.472, ζ = 0.559,%OS = e−ζπ / 1−ζ2x100 = 12.03%, Ts =4ζωn= 1.6 sec, Tp =πωn 1-ζ2= 0.847 sec; ωnTr =1.852, therefore, Tr = 0.414.31.Program:%Form sC(s) to get transfer functionclfnum=[1 3];den=conv([1 3 10],[1 2]);T=tf(num,den)step(T)Computer response:Transfer function:s + 3-----------------------s^3 + 5 s^2 + 16 s + 20
  • 137. 4-30 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.%OS =(0.163 - 0.15)0.15 = 8.67%32.Part c can be approximated as a second-order system. From the exponentially decaying cosine, thepoles are located at s1,2 = −2 ± j9.796 . Thus,Ts =4Re=42= 2 s; Tp =πIm=π9.796= 0.3207 sAlso, ωn = 22+ 9.7962= 10 and ζωn = Re = 2. Hence, ζ = 0.2 , yielding 52.66 percentovershoot.Part d can be approximated as a second-order system. From the exponentially decaying cosine, thepoles are located at S1,2 = −2 ± j9.951. Thus,4 42 s; 0.3157sRe 2 Im 9.951s pT Tπ π= = = = = =Also,2 22 9.951 10.15nω = + = and Re 2nζω = = . Hence, 0.197ζ = , yielding 53.19percent overshoot.33.a.(1) C a 1 s 1s 2 3 s 36+ += =133.7533.75s 1.5+ 2 33.75+= 0.17213 33.75s 1.5+ 2 33.75+=0.17213 5.8095s 1.5+ 2 33.75+
  • 138. Solutions to Problems 4-31Copyright © 2011 by John Wiley & Sons, Inc.Taking the inverse Laplace transformCa1(t) = 0.17213 e-1.5t sin 5.8095t(2) C a 2 s s 2s s 2 3 s 36+ += = 1181s118s 16+s 2 3 s 36+ +− =1181s118s 32+ 0.08333333.7533.75+s 32+233.75+−= 0.055556 1s0.055556 s 32+ 0.014344 33.75+s 32+233.75+−Taking the inverse Laplace transformCa2(t) = 0.055556 - e-1.5t (0.055556 cos 5.809t + 0.014344 sin 5.809t)The total response is found as follows:Cat(t) = Ca1(t) + Ca2(t) = 0.055556 - e-1.5t (0.055556 cos 5.809t - 0.157786 sin 5.809t)Plotting the total response:b.(1) Same as (1) from part (a), or Cb1(t) = Ca1(t)(2) Same as the negative of (2) of part (a), or Cb2 (t) = - Ca2(t)The total response isCbt(t) = Cb1(t) + Cb2(t) = Ca1(t)- Ca2(t) = -0.055556 + e-1.5t (0.055556 cos 5.809t + 0.186474 sin5.809t)
  • 139. 4-32 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.Notice the nonminimum phase behavior for Cbt(t).
  • 140. Solutions to Problems 4-33Copyright © 2011 by John Wiley & Sons, Inc.34.Unit Step2Unit Step1Unit Step1s +3s+102Transfer Fcn21s +3s+102Transfer Fcn11s +3s+102Transfer FcnScope2Scope1ScopeSaturation 20.25 voltsSaturation 10.25 volts10Gain210Gain110GainBacklashDeadzone 0.02
  • 141. 4-34 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.35.sI − A = s1 00 1⎡⎣⎢⎤⎦⎥ −−2 −1−3 −5⎡⎣⎢⎤⎦⎥ =(s + 2) 13 (s + 5)⎡⎣⎢⎤⎦⎥sI − A = s2+ 7s + 7Factoring yields poles at –5.7913 and –1.2087.
  • 142. Solutions to Problems 4-35Copyright © 2011 by John Wiley & Sons, Inc.36.a.sI − A = s1 0 00 1 00 0 1⎡⎣⎢⎢⎤⎦⎥⎥−0 2 30 6 51 4 2⎡⎣⎢⎢⎤⎦⎥⎥=s −2 −30 (s − 6) −5−1 −4 (s − 2)⎡⎣⎢⎢⎤⎦⎥⎥sI − A = s3−8s2−11s + 8b. Factoring yields poles at 9.111, 0.5338, and –1.6448.37.x = (sI - A ) -1 (x0 + B u )123 22 23 22 23 22 21 0 1 2 3 1 130 1 3 1 1 1 93 5 30 54[ 5][ 9]10 12 102[ 5][ 9]( ) [1 2]5 15 54 150( )[ 9][ 5]sss s ss ss s ss sY ss s sY ss s−⎛ ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − + ⋅⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎛ ⎞+ + +⎜ ⎟+ +⎜ ⎟=⎜ ⎟− + −⎜ ⎟⎜ ⎟+ +⎝ ⎠=⎛ ⎞− + −= ⎜ ⎟+ +⎝ ⎠XXX38.x = (sI - A ) -1 (x0 + B u )( ) [0 0 1]Y s = X24 2( )[ 6][ 1][ 0.58579][ 3.4142]s sY ss s s s+ +=+ + + +
  • 143. 4-36 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.39.x = (sI - A ) -1 (x0 + B u )11 0 2 0 3 1 10 1 1 1 0 1ss−⎛ − ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎝ ⎠X3 1[ 2]1 2[ 1][ 2]ss sss s s+⎛ ⎞⎜ ⎟+⎜ ⎟=−⎜ ⎟⎜ ⎟+ +⎝ ⎠X( ) [0 1]Y s = X1 2( )[ 1][ 2]sY ss s s⎛ ⎞−= ⎜ ⎟+ +⎝ ⎠Applying partial fraction decomposition,1 1 3 5 1( )2 1 2 2Y ss s s⎛ ⎞= − +⎜ ⎟+ +⎝ ⎠21 5( ) ( ) 32 2t ty t u t e e− −= − +40.x = (sI − A)−1(x0 + Bu)x = s1 0 00 1 00 0 1⎡⎣⎢⎢⎤⎦⎥⎥−−3 1 00 −6 10 0 −5⎡⎣⎢⎢⎤⎦⎥⎥⎛⎝⎜⎜⎞⎠⎟⎟−1000⎡⎣⎢⎢⎤⎦⎥⎥+011⎡⎣⎢⎢⎤⎦⎥⎥1s⎛⎝⎜⎜⎞⎠⎟⎟x =1s(s + 3)(s + 5)1s(s + 5)1s(s + 5)⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥x(t) =115−16e−3t+110e−5t15−15e−5t15−15e−5t⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥
  • 144. Solutions to Problems 4-37Copyright © 2011 by John Wiley & Sons, Inc.y(t) = 0 1 1[ ]x =25−25e−5t41.Program:A=[-3 1 0;0 -6 1;0 0 -5];B=[0;1;1];C=[0 1 1];D=0;S=ss(A,B,C,D)step(S)Computer response:a =x1 x2 x3x1 -3 1 0x2 0 -6 1x3 0 0 -5b =u1x1 0x2 1x3 1c =x1 x2 x3y1 0 1 1d =u1y1 0Continuous-time model.
  • 145. 4-38 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.42.Program:syms s %Construct symbolic object for%frequency variable s.a %Display labelA=[-3 1 0;0 -6 1;0 0 -5] %Create matrix A.B=[0;1;1]; %Create vector B.C=[0 1 1]; %Create C vectorX0=[1;1;0] %Create initial condition vector,X(0).U=1/s; %Create U(s).I=[1 0 0;0 1 0;0 0 1]; %Create identity matrix.X=((s*I-A)^-1)*(X0+B*U); %Find Laplace transform of state vector.x1=ilaplace(X(1)) %Solve for X1(t).x2=ilaplace(X(2)) %Solve for X2(t).x3=ilaplace(X(3)) %Solve for X3(t).y=C*[x1;x2;x3] %Solve for output, y(t).y=simplify(y) %Simplify y(t).y(t) %Display label.pretty(y) %Pretty print y(t).Computer response:ans =aA =-3 1 00 -6 10 0 -5X0 =110x1 =7/6*exp(-3*t)-1/3*exp(-6*t)+1/15+1/10*exp(-5*t)x2 =exp(-6*t)+1/5-1/5*exp(-5*t)x3 =1/5-1/5*exp(-5*t)y =2/5+exp(-6*t)-2/5*exp(-5*t)y =2/5+exp(-6*t)-2/5*exp(-5*t)ans =y(t)2/5 + exp(-6 t) - 2/5 exp(-5 t)43.|λI - A | = λ2 + 5λ +1|λI - A | = (λ + 0.20871) (λ + 4.7913)
  • 146. Solutions to Problems 4-39Copyright © 2011 by John Wiley & Sons, Inc.Therefore,Solving for Ais two at a time, and substituting into the state-transition matrixTo find x(t),To find the output,44.|λI - A | = λ2 + 1
  • 147. 4-40 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.Solving for the Ais and substituting into the state-transition matrix,To find the state vector,(3,4)1 cos[ ](3,4)sin[ ]( 3cos[ ] 4sin[ ] 3)y xtyty t t=−⎛ ⎞Δ = ⎜ ⎟⎝ ⎠= − + +45.|λI - A | = (λ + 2) (λ + 0.5 - 2.3979i) (λ + 0.5 + 2.3979i)Let the state-transition matrix be
  • 148. Solutions to Problems 4-41Copyright © 2011 by John Wiley & Sons, Inc.Since φ(0) = I, Φ.(0) = A, and φ..(0) = A2, we can evaluate the coefficients, Ais. Thus,Solving for the Ais taking three equations at a time,U s in g x (t) = φ (t)x (0 ) + ∫0tφ (t-τ )B u (τ )dτ , an d y = 1 0 0 x (t),=12 -12 e-2t46.Program:syms s t tau %Construct symbolic object for%frequency variable s, t, and tau.a %Display label.A=[-2 1 0;0 0 1;0 -6 -1] %Create matrix A.B=[1;0;0] %Create vector B.C=[1 0 0] %Create vector C.X0=[1;1;0] %Create initial condition vector,X(0).I=[1 0 0;0 1 0;0 0 1]; %Create identity matrix.E=(s*I-A)^-1 %Display label.E=((s*I-A)^-1) %Find Laplace transform of state%transition matrix, (sI-A)^-1.Fi11=ilaplace(E(1,1)); %Take inverse Laplace transformFi12=ilaplace(E(1,2)); %of each elementFi13=ilaplace(E(1,3));Fi21=ilaplace(E(2,1));Fi22=ilaplace(E(2,2));Fi23=ilaplace(E(2,3));
  • 149. 4-42 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.Fi31=ilaplace(E(3,1));Fi32=ilaplace(E(3,2)); %to find state transition matrix.Fi33=ilaplace(E(3,3)); %of (sI-A)^-1.Fi(t) %Display label.Fi=[Fi11 Fi12 Fi13 %Form Fi(t).Fi21 Fi22 Fi23Fi31 Fi32 Fi33];pretty(Fi) %Pretty print state transition matrix, Fi.Fitmtau=subs(Fi,t,t-tau); %Form Fi(t-tau).Fi(t-tau) %Display label.pretty(Fitmtau) %Pretty print Fi(t-tau).x=Fi*X0+int(Fitmtau*B*1,tau,0,t);%Solve for x(t).x=simple(x); %Collect terms.x=simplify(x); %Simplify x(t).x=vpa(x,3);x(t) %Display label.pretty(x) %Pretty print x(t).y=C*x; %Find y(t)y=simplify(y);y=vpa(simple(y),3);y=collect(y);y(t)pretty(y) %Pretty print y(t).Computer response:ans =aA =-2 1 00 0 10 -6 -1B =100C =1 0 0X0 =110ans =E=(s*I-A)^-1E =[ 1/(s+2), (s+1)/(s+2)/(s^2+s+6), 1/(s+2)/(s^2+s+6)][ 0, (s+1)/(s^2+s+6), 1/(s^2+s+6)][ 0, -6/(s^2+s+6), s/(s^2+s+6)]
  • 150. Solutions to Problems 4-43Copyright © 2011 by John Wiley & Sons, Inc.ans =Fi(t)[ 13[exp(-2 t) , - 1/8 exp(-2 t) + 1/8 %1 + --- %2 ,[ 184]1/8 exp(-2 t) - 1/8 %1 + 3/184 %2]][[0 , 1/23 %2 + %1 , - 1/231/2 1/2 1/2(-23) (exp((-1/2 + 1/2 (-23) ) t) - exp((-1/2 - 1/2 (-23) ) t))]][[0 , 6/231/2 1/2 1/2(-23) (exp((-1/2 + 1/2 (-23) ) t) - exp((-1/2 - 1/2 (-23) ) t))], - 1/23 %2 + %1]1/2%1 := exp(- 1/2 t) cos(1/2 23 t)1/2 1/2%2 := exp(- 1/2 t) 23 sin(1/2 23 t)ans =Fi(t-tau)[[exp(-2 t + 2 tau) ,[13 1/2- 1/8 exp(-2 t + 2 tau) + 1/8 %2 cos(%1) + --- %2 23 sin(%1) ,1841/2 ]1/8 exp(-2 t + 2 tau) - 1/8 %2 cos(%1) + 3/184 %2 23 sin(%1)]][ 1/2 1/2[0 , 1/23 %2 23 sin(%1) + %2 cos(%1) , - 1/23 (-23) (1/2exp((-1/2 + 1/2 (-23) ) (t - tau))1/2 ]- exp((-1/2 - 1/2 (-23) ) (t - tau)))][ 1/2 1/2[0 , 6/23 (-23) (exp((-1/2 + 1/2 (-23) ) (t - tau))1/2- exp((-1/2 - 1/2 (-23) ) (t - tau))) ,
  • 151. 4-44 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.1/2 ]- 1/23 %2 23 sin(%1) + %2 cos(%1)]1/2%1 := 1/2 23 (t - tau)%2 := exp(- 1/2 t + 1/2 tau)ans =x(t)[.375 exp(-2. t) + .125 exp(-.500 t) cos(2.40 t)+ .339 exp(-.500 t) sin(2.40 t) + .500][.209 exp(-.500 t) sin(2.40 t) + exp(-.500 t) cos(2.40 t)][1.25 i (exp((-.500 + 2.40 i) t) - 1. exp((-.500 - 2.40 i) t))]ans =y(t).375 exp(-2. t) + .125 exp(-.500 t) cos(2.40 t)+ .339 exp(-.500 t) sin(2.40 t) + .50047.The state-space representation used to obtain the plot is,x.=0 1-1 -0.8x +01u(t); y(t) = 1 0 xUsing the Step Response software,
  • 152. Solutions to Problems 4-45Copyright © 2011 by John Wiley & Sons, Inc.Calculating % overshoot, settling time, and peak time,2ζωn = 0.8, ωn = 1, ζ = 0.4. Therefore, %OS = e−ζπ / 1−ζ2x100 = 25.38%, Ts =4ζωn= 10 sec,Tp =πωn 1-ζ2= 3.43 sec.48.
  • 153. 4-46 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.49.a. P(s) =s+0.5s(s+2)(s+5) =1/20s +1/4s+2 -3/10s+5 . Therefore, p(t) =120 +14 e-2t -310 e-5t.b. To represent the system in state space, draw the following block diagram.1s2+7s+10s+0.5V(s) Y(s) P(s)For the first block,y..+ 7y.+10y = v(t)Let x1 = y, and x2 = y.. Therefore,x.1 = x2x.2 = -10x1 - 7x2 + v(t)Also,p(t) = 0.5y + y.= 0.5x1 + x2Thus,x.=0 1-10 -7x +011; p(t) = 0.5 1 xc.Program:A=[0 1;-10 -7];B=[0;1];C=[.5 1];D=0;S=ss(A,B,C,D)step(S)
  • 154. Solutions to Problems 4-47Copyright © 2011 by John Wiley & Sons, Inc.Computer response:a =x1 x2x1 0 1x2 -10 -7b =u1x1 0x2 1c =x1 x2y1 0.5 1d =u1y1 0Continuous-time model.50.a. ωn = 10 = 3.16; 2ζωn = 4. Therefore ζ = 0.632. %OS = e− ξπ / 1−ξ2*100 = 7.69%.Ts =4ξωn= 2 seconds. Tp =πωn 1− ξ 2= 1.28 seconds. From Figure 4.16, Trωn = 1.93.Thus, Tr = 0.611 second. To justify second-order assumption, we see that the dominant poles are at –2 ± j2.449. The third pole is at -10, or 5 times further. The second-order approximation is valid.
  • 155. 4-48 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.b. Ge(s) =K(s+10)(s2+4s+10)=Ks3+14s2+50s+100. Representing the system in phase-variable form:A =0 1 00 0 1−100 −50 −14⎡⎣⎢⎢⎤⎦⎥⎥; B =00K⎡⎣⎢⎢⎤⎦⎥⎥; C = 1 0 0[ ]c.Program:numg=100;deng=conv([1 10],[1 4 10]);G=tf(numg,deng)step(G)Computer response:Transfer function:100-------------------------s^3 + 14 s^2 + 50 s + 100%OS =(1.08-1)1 * 100 = 8%
  • 156. Solutions to Problems 4-49Copyright © 2011 by John Wiley & Sons, Inc.51.a. ωn = 0.28 = 0.529; 2ζωn = 1.15. Therefore ζ = 1.087.b. P(s) = U(s)7.63x10-2s2+1.15s+0.28, where U(s) =2s . Expanding by partial fractions, P(s) =0.545s +natural response terms. Thus percent paralysis = 54.5%.c. P(s) =7.63x10-2s(s2+1.15s+0.28)=0.2725s -0.48444s+0.35 +0.21194s+0.8 .Hence, p(t) = 0.2725 - 0.48444e-0.35t + 0.21194e-0.8t. Plotting,Fractionalparalysisfor1%isofluraned. P(s) =Ks *7.63x10-2s2+1.15s+0.28=1s + natural response terms. Therefore,7.63x10-2 K0.28 = 1. Solvingfor K, K = 3.67%.52.a. Writing the differential equation,dc(t)dt = -k10c(t) +i(t)VdTaking the Laplace transform and rearranging,(s+k10)C(s) =I(s)Vdfrom which the transfer function is found to beC(s)I(s) =1Vds+k10For a step input, I(s) =I0s . Thus the response isC(s) =I0Vds(s+k10) =I0k10Vd(1s -1s+k10)Taking the inverse Laplace transform,
  • 157. 4-50 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.c(t) =I0k10Vd(1− e−k10 t)where the steady-state value, CD, isCD =I0k10VdSolving for IR = I0,IR = CDk10Vdb. Tr =2.2k10; Ts =4k10c. IR = CDk10Vd = 12μgml x 0.07 hr-1 x 0.6 liters = 0.504mghd. Using the equations of part b, where k10 = 0.07, Tr = 31.43 hrs, and Ts = 57.14 hrs.53. Consider the un-shifted Laplace transform of the output2222)20()286.14()125)(814.5(82.280)05.01()07.01()008.01)(172.01(5.2)(++++=++++=sssssssssssY)20()20()286.14()286.14( 22++++++++=sEsDsCsBsA5.2)20()286.14()125)(814.5(82.280022=++++= =sssssA7.564)20()125)(814.5(82.280286.142=+++= −=sssssB286.14232286.1424004094.2040852.3673582.280)20()125)(814.5(82.280−=−=++++=+++= sssssssdsdssssdsdC2232223)40040()400803)(94.2040852.3673582.280()2.3673564.561)(40040(−=++++++−+++= ssssssssssss7.219−=57.640)286.14()125)(814.5(82.280202=+++= −=sssssD2023220209.204572.2894.2040852.3673582.280)286.14()125)(814.5(82.280−=−=++++=+++= sssssssdsdssssdsdE
  • 158. Solutions to Problems 4-51Copyright © 2011 by John Wiley & Sons, Inc.202232223)09.204572.28()09.204144.573)(94.2040852.3673582.280()2.3673564.561)(09.204572.28(−=++++++−+++= ssssssssssss18.217=thus)20(18.217)20(57.640)286.14(7.219)286.14(7.5645.2)( 22+++++−++=ssssssYObtaining the inverse Laplace transform of the latter and delaying the equation in time domain weget)008.0(]18.217)008.0(57.6407.219)008.0(7.5645.2[)()008.0(20)008.0(20)008.0(286.14)008.0(286.14−+−+−−+=−−−−−−−−tueeteettytttt54.a. The transfer function can be written as)44.172.0)(1()33.3(5056.2)( 21.0++++=−sssessIsθIt has poles at s=-0.36±j1.145 and s=-1. A zero at s=-3.33The ‘far away’ pole at -1 is relatively close to the complex conjugate poles as 0.36*5>1 so adominant pole approximation can’t be applied.b) In time domain the input can be expressed as:))15.0()((250)( −−= tutuAti μObtaining Laplace transforms this can be expressed as
  • 159. 4-52 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.sesIs15.01250)(−−= μWe first obtain the response to an unshifted unit step:44.172.01)44.172.0)(1()33.3(5056.2)( 22++++++=++++=ssDCssBsAssssssθ8.5)44.172.0)(1()33.3(5056.202=++++= =sssssA4.3)72.1)(1()33.2(5056.2)44.172.0()33.3(5056.212−=−=+++= −=sssssBWe will get C and D by equating coefficients. Substituting these two values and multiplying bothsides by the denominator we get.)1()()44.172.0(4.3)44.172.0)(1(8.5)33.3(5056.2 22+++++−+++=+ ssDCssssssss3 2 3 23 22.5056( 3.33) 5.8( 1.72 2.16 1.44) 3.4( 0.72 1.44 )( ( ) )s s s s s s sCs C D s Ds+ = + + + − + ++ + + +352.8)632.10()528.7()4.2()33.3(5056.2 23+++++++=+ sDsDCsCsWe immediately get C=-2.4 and D=-5.128So3104.1)36.0(128.54.214.38.544.172.0128.54.214.38.5)( 22+++−+−=+++−+−=ssssssssssθ
  • 160. Solutions to Problems 4-53Copyright © 2011 by John Wiley & Sons, Inc.2 225.8 3.4 2.4( 0.15) 4.768 5.8 3.4 2.4( 0.15)1 ( 0.36) 1.3104 1 ( 0.36) 1.31041.1454.164( 0.36) 1.3104s ss s s s s ss+ + += − − = − −+ + + + + +−+ +Obtaining inverse Laplace transform we get)145.1sin(164.4)145.1cos(4.24.38.5)( 36.036.0teteet ttt −−−−−−=θ)30145.1sin(4.24.38.5 36.0 o+−−= −−tee ttSo the actual (shifted) unit step response is given by)1.0()]30)1.0(145.1sin(4.24.38.5[)( )1.0(36.0)1.0(−+−−−= −−−−tuteet tt oθThe response to the pulse is given by:−−+−−−= −−−−)1.0()]30)1.0(145.1sin(6.085.045.1[)( )1.0(36.0)1.0(tutmememt tt oθ)25.0()]30)25.0(145.1sin(6.085.045.1[ )25.0(36.0)25.0(−+−−− −−−−tutmemem tt o55.At steady state the input is ≈ 9V and the output is ≈ 6V Thus G(0)=6/9=0.667The maximum peak is achieved at ≈ 285μ with a %OS = (7.5/6-1)*100 = 25%This corresponds to a damping factor of4.09218.13863.1)100/(%ln)100/ln(%222≈+=+−=ππζOSOS2.12027)9165.0)(285(1 2==−=μπζπωpnTSo the approximated transfer function is
  • 161. 4-54 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.72622222210*5.14962210*5.962.120272.12027*4.0*22.12027*667.02)(++=++=++=ssssssKsGnnnωζωω56.The oscillation period is212ζωπ−= nTand from the figure sssT0169.00506.00675.02=−=Thus T=0.0338sec from which we get 8931.1851 2=−ζωnThe peaks of the response occur when the ‘cos’ term of the step response is ±1 thus from the figurewe have:1492.1112)0506.0(=−+−ζζωneand 9215.0112)0675.0(=−−−ζζωneFrom which we get9006.10785.01492.0)0675.0()0506.0(==−−nneeζωζωor 9006.1)0169.0(=− ne ζωor 38=nζωSubstituting this result we get 8931.1851381 22=−=− ζζζωnor 2284.34556)1(1444 22=−ζζor 0436.02=ζ or 21.0=ζFinally 9.18038==ζωn57.The step input amplitude is the same for both responses so it will just be assumed to be unitary.For the ‘control’ response we have:018.0=finalc , 024.0=ptM from which we get%33.33%100018.0018.0024.0%100%max=×−=×−=finalfinalcccOS33.0)333.0(ln)333.0ln()100/(%ln)100/ln(%2222=+−=+−=ππζOSOS3.33333.011.01 22=−=−=πζπωpnT
  • 162. Solutions to Problems 4-55Copyright © 2011 by John Wiley & Sons, Inc.Leading a transfer function9.1108229.1108)( 2++=sssGcSimilarly for the ‘hot tail’:023.0=finalc , 029.0=ptM%1.26%100023.0023.0029.0% =×−=OS393.0)261.0(ln)261.0ln(22=+−=πζ17.34261.011.01 22=−=−=πζπωpnT6.11679.266.1167)( 2++=sssGhUsing MATLAB:>> syms s>> s=tf(s)Transfer function:s>> Gc = 1108.89/(s^2+22*s+1108.89);>> Gh = 1167.6/(s^2+26.9*s+1167.6);>> step(Gc,Gh)
  • 163. 4-56 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.0 0.1 0.2 0.3 0.4 0.5 0.600.20.40.60.811.21.4Step ResponseTime (sec)AmplitudeBoth responses are equivalent if error tolerances are considered.58.The original transfer function has zeros at 74007200 js ±−=And poles at 45001900 js ±−= ; 1520120 js ±−=With 1864.0)0( =GThe dominant poles are those with real parts at -120, so a real pole is added at-1200 giving the following approximation:)1200)(108.2324240()106.10614400(106.106)108.2324)(1200(1864.0)( 326263+×++×+−××≈ssssssG)1200)(108.2324240()106.10614400(8782.43262+×++×+−=sssssUsing MATLAB:>> syms s>> s=tf(s);
  • 164. Solutions to Problems 4-57Copyright © 2011 by John Wiley & Sons, Inc.>>G=9.7e4*(s^2-14400*s+106.6e6)…/(s^2+3800*s+23.86e6)/(s^2+240*s+2324.8e3);>> Gdp=4.8782*(s^2-14400*s+106.6e6)/(s^2+240*s+2324.8e3)/(s+1200);>> step(G,Gdp)0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.0500.050.10.150.20.250.30.35Step ResponseTime (sec)AmplitudeBoth responses differ because the original non-dominant poles are very close to the complex pair ofzeros.59.M(s) requires at least 4 ‘far away’ poles that are added a decade beyond all original poles and zeros.This gives4224822)1.0)(0001.0()0001.0018.0()009.0(81.1028)1.0/1)(0001.0(1072.9)0001.0018.0()009.0()(+++++=++×+++= −sssssssssssM
  • 165. 4-58 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.60.36.0)30.0(ln)30.0ln()100/(%ln)100/ln(%2222=+−=+−=ππζOSOS026.030.011271 22=−=−=πζπωpnT00067.00187.000067.02)( 2222++=++=sssssGnnnωζωω61.a. Let the impulse response of T(s) be h(t). We have that205)20)(5(450)(+++=++=sBsAsssH30204505 =+= −=ssA ; 30545020 −=+= −=ssB2030530)(+−+=sssH . Obtaining the inverse Laplace transform we gettteeth 2053030)( −−−=b.Let the step response of the system be g(t). We have thatttttt tttteedtedtedtthtg 020050 0205020305303030)()( −−−−−−−=−== ∫ ∫∫tttteeee 2052055.165.4)1(5.1)1(6 −−−−+−=−+−−=c.205)20)(5(450)(++++=++=sCsBsAssssG5.4)20)(5(4500 =++= =sssA ; 6)20(4505 −=+= −=sssB ; 5.1)5(45020 =+= −=sssCLeading205.1565.4)(+++−=ssssG . After the inverse Laplace we gettteetg 2055.165.4)( −−+−=62.a.The poles given by 01097.31099.8 332=×+×+ −−ss have an sec/063.0 radn =ω and0714.0=ζ
  • 166. Solutions to Problems 4-59Copyright © 2011 by John Wiley & Sons, Inc.The poles given by 023.1821.42=++ ss have an sec/27.4 radn =ω and 493.0=ζ Thusthe former represent the Phugoid and the latter the Short Period modes.b.In the original we have 85.4)0( −=eδθso the Phugoid approximation is given by:)1097.31099.8()0098.0(965.1332 −−×+×++−≈ssseδθc.>> syms s>> s=tf(s);>>G=-26.12*(s+0.0098)*(s+1.371)/(s^2+8.99e-3*s+3.97e-3)/(s^2+4.21*s+18.23);>> Gphug=-1.965*(s+0.0098)/(s^2+8.99e-3*s+3.97e-3);>> step(G,Gphug)0 500 1000 1500-40-30-20-1001020Step ResponseTime (sec)AmplitudeBoth responses are indistinguishable.
  • 167. 4-60 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.63.a.Programnumg=[33 202 10061 24332 170704];deng=[1 8 464 2411 52899 167829 913599 1076555];G=tf(numg,deng)[K,p,k]=residue(numg,deng)Computer ResponseK =0.0018 + 0.0020i0.0018 - 0.0020i-0.1155 - 0.0062i-0.1155 + 0.0062i0.0077 - 0.0108i0.0077 + 0.0108i0.2119p =-1.6971 +16.4799i-1.6971 -16.4799i-0.5992 +12.1443i-0.5992 -12.1443i-1.0117 + 4.2600i-1.0117 - 4.2600i-1.3839k =[]b.Therefore, an approximation to G(s)/ is:0.21191.3839( )sG s+=c.Programnumg=[33 202 10061 24332 170704];deng=[1 8 464 2411 52899 167829 913599 1076555];G=tf(numg,deng);numga=0.2119;denga=[1 1.3839];
  • 168. Solutions to Problems 4-61Copyright © 2011 by John Wiley & Sons, Inc.Ga=tf(numga,denga);step(G)hold onstep(Ga)Computer ResponseApproximation does not show oscillations and is slightly off of final value.64.Computer ResponseTransfer function:s^15 + 1775 s^14 + 1.104e006 s^13 + 2.756e008 s^12 + 2.272e010 s^11+ 7.933e011 s^10 + 1.182e013 s^9 + 6.046e013 s^8 + 1.322e014 s^7+ 1.238e014 s^6 + 3.977e013 s^5 + 5.448e012 s^4 + 3.165e011 s^3+ 6.069e009 s^2 + 4.666e007 s + 1.259e005----------------------------------------------------------------------------31.62 s^17 + 4.397e004 s^16 + 1.929e007 s^15 + 2.941e009 s^14+ 1.768e011 s^13 + 4.642e012 s^12 + 5.318e013 s^11 + 2.784e014 s^10+ 7.557e014 s^9 + 1.238e015 s^8 + 1.356e015 s^7 + 8.985e014 s^6+ 2.523e014 s^5 + 3.179e013 s^4 + 1.732e012 s^3 + 3.225e010 s^2+ 2.425e008 s + 6.414e005
  • 169. 4-62 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.65.a. To find the step responses for these two processes, ya(t) and yp(t), we consider first the un-shiftedLaplace transform of their outputs for Xa(s) = Xp(s) = 1/s:)1077.6()1077.6(108.9)126.1478(49.14)( 443*−−−×++=×+×=+=sBsAsssssYa (1),where 49.1401077.6108.943==×+×= −−ssA and49.141077.6108.943−=×−=×=−−ssB (2)Substituting the values of A and B into equation (1) gives:⎟⎟⎠⎞⎜⎜⎝⎛×+−=×++= −−)1076.6(1149.14)1076.6()( 44*sssBsAsYa (3)Taking the inverse Laplace transform of )(*sYa and delaying the resulting response in the timedomain by 4 seconds, we get:)4(])4(1076.61[49.14)(4−−×−−=−tutetya (4)
  • 170. Solutions to Problems 4-63Copyright © 2011 by John Wiley & Sons, Inc.Noting that the denominator of )(sGp can be factored into)10814.6)(10174.0( 33 −−×+×+ ss , we have:)10814.6()10174.0()10814.6)(10174.0(10716.1)( 33335*−−−−−×++×++=×+×+×=sEsDsCssssYp (5),where: 48.140)10814.6)(10174.0(10716.1335==×+×+×= −−−sssC ;85.1410174.0)10814.6(10716.1335−=×−=×+×=−−−sssD ;37.010814.6)10174.0(10716.1335=×−=×+×=−−−sssE . (6)Substituting the values of C, D and E into equation (5) and simplifying gives:⎟⎟⎠⎞⎜⎜⎝⎛×++×+−=×++×+−= −−−−)10814.6(0256.0)10174.0(0256.1148.14)10814.6(37.0)10174.0(85.1448.14)( 3333*sssssssYp (7)Taking the inverse Laplace transform of )(*sYp and delaying the resulting response in the timedomain by 30 seconds, we get:)30(])30(10814.60256.0)30(10174.00256.11[48.14)(33−−×−+−×−−=−−tutetetyp (8)b.Using Simulink to model the two processes described above, ya(t) and yp(t) were output to the“workspace.” Matlab plot commands were then utilized to plot ya(t) and yp(t) on a single graph,which is shown below.
  • 171. 4-64 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.XY GraphTransportDelayTransfer Fcn1478 .26s+114.49To Workspace2pHTo Workspace1timeStepClockXY GraphTransportDelayTransfer Fcns +6.989 E-3s+1.185 E-621.716 E-5To Workspace2pH2To Workspace1timeStepClock
  • 172. Solutions to Problems 4-65Copyright © 2011 by John Wiley & Sons, Inc.
  • 173. 4-66 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.66.a.>> A=[-8.792e-3 0.56e-3 -1e-3 -13.79e-3; -0.347e-3 -11.7e-3 -0.347e-3 0; 0.261 -20.8e-3 -96.6e-30; 0 0 1 0]A =-0.0088 0.0006 -0.0010 -0.0138-0.0003 -0.0117 -0.0003 00.2610 -0.0208 -0.0966 00 0 1.0000 0>> eig(A)ans =-0.19470.0447 + 0.1284i0.0447 - 0.1284i-0.0117b.Given the eigenvalues, the state-transition matrix will be of the form⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=44434241343332312423222114131211)(KKKKKKKKKKKKKKKKtΦ with))1284.0cos()1284.0sin( 0447.040447.030117.021947.01 teKteKeKeKK tijtijtijtijij++−−+++=Thus 64 constants have to be found.
  • 174. Solutions to Problems 4-67Copyright © 2011 by John Wiley & Sons, Inc.67.a. The equations are rewritten assCLELuLddtdi 11+−−=CLCuRCiCddtdu 11−−=from which we obtainsCLCLELuiRCCdLddtdudtdi⎥⎥⎦⎤⎢⎢⎣⎡+⎥⎦⎤⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡011110[ ] ⎥⎦⎤⎢⎣⎡=CLiiy 10b.To obtain the transfer function we first calculateLCdRCsssCdLdRCsRCsCdLdss 211)1()1(111111)(−++⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−+=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+−−−=−−−AISo[ ]⎥⎥⎦⎤⎢⎢⎣⎡−++⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−+=−= −01)1()1(11110)()( 21LLCdRCsssCdLdRCsssG BAICLCdsRCsLCdLLCdsRCssCd2222 )1(1101)1(11−++−=⎥⎥⎦⎤⎢⎢⎣⎡−++⎥⎦⎤⎢⎣⎡ −=
  • 175. 4-68 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.68.a.We have ⎥⎦⎤⎢⎣⎡−+=−sss126.234.8)( AI and)06.8)(28.0(34.8126.226.234.834.8126.2)( 21++⎥⎦⎤⎢⎣⎡+−=++⎥⎦⎤⎢⎣⎡+−=− −sssssssss AIWe first find⎭⎬⎫⎩⎨⎧++ )06.8)(28.0(1ss1-L06.828.0)06.8)(28.0(1 21+++=++ sKsKss129.006.8128.01 =+= −=ssK ; 129.028.0106.82 −=+= −=ssK sottee).)(s.(s06.828.0129.0129.00682801 −−−=⎭⎬⎫⎩⎨⎧++1-LFollows thattteess.ss06.828.0292.0292.0)06.8)(28.0(1262)06.8)(28.0(26.2 −−+−=⎭⎬⎫⎩⎨⎧++=⎭⎬⎫⎩⎨⎧++− 1-1-L-Ltteessdtdsss 06.828.004.1036.0)06.8)(28.0(1)06.8)(28.0(−−+−=⎭⎬⎫⎩⎨⎧++=⎭⎬⎫⎩⎨⎧++1-1-LLAnd⎭⎬⎫⎩⎨⎧+++⎭⎬⎫⎩⎨⎧++=⎭⎬⎫⎩⎨⎧+++)06.8)(28.0(1348)06.8)(28.0()06.8)(28.0(34.8ss.ssssss 1-1-1-LLLtttttteeeeee 06.828.006.828.006.828.0036.004.1076.1076.104.1036.0 −−−−−−−=−++−=Finally the state transition matrix is given by:⎥⎦⎤⎢⎣⎡−−+−+−= −−−−−−−−tttttttteeeeeeeet 06.828.006.828.006.828.006.828.0036.004.1129.0129.0292.0292.004.1036.0)(Φ
  • 176. Solutions to Problems 4-69Copyright © 2011 by John Wiley & Sons, Inc.b.⎥⎦⎤⎢⎣⎡−+−= −−−−tttteeeet 06.828.006.828.0129.0129.004.1036.0)( BΦtttttteeeeeet 06.828.006.828.006.828.0748.12159.0292.0292.004.13451.0)( −−−−−−+−=−++−=BCΦSince 1)( =tu∫∫−−−−+−=−=ttttdeedtty0)(06.8)(28.00]748.12159.0[)()( τττ ττBCΦ∫ ∫−−+−=t tttdeedee0 006.806.828.028.0748.12159.0 ττ ττttteeee 006.806.828.028.006.8748.1228.0159.0 ττ −−+−=tttteeee 06.828.006.828.0582.1568.0014.1]1[582.1]1[568.0 −−−−−+=−+−−=c.>> A=[-8.34 -2.26; 1 0];>> B = [1; 0];>> C = [12.54 2.26];>> D = 0;>> t = linspace(0,15,1000);>> y1 = step(A,B,C,D,1,t);>> y2 = 1.014+0.568*exp(-0.28.*t)-1.582*exp(-8.06.*t);>> plot(t,y1,t,y2)
  • 177. 4-70 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.0 5 10 15-0.200.20.40.60.811.21.41.6SOLUTIONS TO DESIGN PROBLEMS69.Writing the equation of motion, ( fvs + 2)X(s) = F(s). Thus, the transfer function isX(s)F(s)=1/ fvs +2fv. Hence, Ts =4a=42fv= 2 fv , or fv =Ts2.70.The transfer function is, F(s) =1/ Ms2+1Ms +KM. Now,4 44 81Re2sT MM= = = = . Thus,12M = . Substituting the value of M in the denominator of the transfer function yields,22 2s s K+ + . Identify the roots 1,2 1 2 1s j K= − ± − . Using the imaginary part and substitutinginto the peak time equation yields 1Im 2 1pTKπ π= = =−, from which 5.43K = .
  • 178. Solutions to Design Problems 4-71Copyright © 2011 by John Wiley & Sons, Inc.71.Writing the equation of motion, (Ms2+ fvs +1)X(s) = F(s) . Thus, the transfer function isX(s)F(s)=1/ Ms2+fvMs +1M. Since Ts =10 =4ζωn, ζωn = 0.4 . But,fvM= 2ζωn = 0.8.Also,from Eq. (4.39) 17% overshoot implies ζ = 0.491. Hence, ωn = 0.815. Now, 1/M = ωn2= 0.664.Therefore, M 1.51. Since 2 0.8, 1.21vn vffMζω= = = .72.Writing the equation of motion: (Js2+s+K)θ(s) = T(s). Therefore the transfer function isθ(s)T(s) =1Js2+1Js+KJ.ζ =- ln (%OS100 )π2 + ln2 (%OS100 )= 0.358.Ts =4ζωn=412J= 8J = 3.Therefore J =38. Also, Ts = 3 =4ζωn=4(0.358)ωn. Hence, ωn = 3.724. Now,KJ = ωn2 = 13.868.Finally, K = 5.2.73.Writing the equation of motion[s2+D(5)2s+14 (10) 2]θ(s) = T(s)The transfer function isθ (s)T(s) =1s2+25Ds+25Also,ζ =- ln (%OS100 )π2 + ln2 (%OS100 )= 0.358and2ζωn = 2(0.358)(5) = 25DTherefore D = 0.14.
  • 179. 4-72 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.74.The equivalent circuit is:where Jeq = 1+(N1N2)2 ; Deq = (N1N2)2; Keq = (N1N2)2. Thus,θ1(s)T(s) =1Jeqs2+Deqs+Keq. LettingN1N2= n and substituting the above values into the transferfunction,θ1(s)T(s) =11+n2s2 +n21+n2 s +n21+n2. Therefore, ζωn =n22(1+n2). Finally, Ts =4ζωn=8(1+n2)n2 = 16. Thusn = 1.75.Let the rotation of the shaft with gear N2 be θL(s). Assuming that all rotating load has been reflectedto the N2 shaft, JeqLs2+ DeqLs + K( )θL (s) + F(s)r = Teq (s) , where F(s) is the force from thetranslational system, r = 2 is the radius of the rotational member, JeqL is the equivalent inertia at theN2 shaft, and DeqL is the equivalent damping at the N2 shaft. Since JeqL = 1(2)2+ 1 = 5 and DeqL =1(2)2= 4, the equation of motion becomes, 5s2+ 4s + K( )θL (s) + 2F(s) = Teq (s). For thetranslational system (Ms2+ s)X(s) = F(s) . Substituting F(s) into the rotational equation ofmotion, 5s2+ 4s + K( )θL (s) + Ms2+ s( )2X(s) = Teq (s).But,θL (s) =X(s)r=X(s)2and Teq (s) = 2T(s). Substituting these quantities in the equationabove yields (5 + 4M)s2+ 8s + K( )X(s)4= T(s). Thus, the transfer function isX(s)T(s)=4 /(5 + 4M)s2+8(5 + 4M)s +K(5 + 4M). Now,4 415 (5 4 )8Re2(5 4 )sT MM= = = = ++.Hence, M = 5/2. For 10% overshoot, ζ = 0.5912 from Eq. (4.39). Hence,
  • 180. Solutions to Design Problems 4-73Copyright © 2011 by John Wiley & Sons, Inc.82 0.5333(5 4 )nMζω = =+. Solving for ωn yields ωn = 0.4510. But,0.4510.(5 4 ) 15nK KMω = = =+Thus, K = 3.051.76.The transfer function for the capacitor voltage isVC(s)V(s) =1CsR+Ls+1Cs=106s2+Rs+106 .For 20% overshoot, ζ =- ln (%OS100 )π2 + ln2 (%OS100 )= 0.456. Therefore, 2ζωn = R = 2(0.456)(103) =912Ω.77.Solving for the capacitor voltage using voltage division, VC (s) = Vi (s)1/(CS)R + LS +1CS. Thus, thetransfer function isVC (s)Vi (s)=1/(LC)s2+RLs +1LC. Since Ts =4Re=10−3, Re =R2L= 4000. ThusR = 8 KΩ. Also, since 20% overshoot implies a damping ratio of 0.46 and2ζωn = 8000, ωn = 8695.65 =1LC. Hence, C = 0.013 μF.78.Using voltage division the transfer function is,VC (s)Vi (s)=1CsR+ Ls +1Cs=1LCs2+RLs +1LCAlso,3 4 4 87 10Re2sLT xR RL−= = = = . Thus, 1143RL= . Using Eq. (4.39) with 15% overshoot, ζ= 0.5169. But, 2ζωn = R/L. Thus, 51 11105.63(10 )nLC Lω −= = = . Therefore, L = 81.8 mHand R = 98.5 Ω.
  • 181. 4-74 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.79.For the circuit shown belowR1 =L =i1(t) i2(t)owrite the loop equations asR 1 L s+ I 1 s R 1 I 2 s− V i s=R 1 I 1 s− R 1 R 21C s+ + I 2 s+ 0=Solving for I2(s)I 2 sR 1 L s+ V i sR 1− 0R 1 L s+ R 1−R 1− R 1 R 21C s+ +=( )But, V o s 1C sI 2 s= . Thus,V o sV i sR 1R 2 R 1+ C L s 2 C R 2 R 1 L+ s R 1+ +=Substituting component values,Vo (s)Vi (s)= 10000001(R2 + 1000000)Cs2+(1000000CR2 +1)(R2 +1000000)Cs +10000001(R2 +1000000)CFor 8% overshoot, ζ = 0.6266. For Ts = 0.001, ζωn =40.001 = 4000. Hence, ωn = 6383.66. Thus,2211000000 6383.66( 1000000)R C=+or,210.02451000000CR=+(1)Also,
  • 182. Solutions to Design Problems 4-75Copyright © 2011 by John Wiley & Sons, Inc.1000000 C R 2 1+R 2 1000000+ C8000= (2)Solving (1) and (2) simultaneously, 2 8023R = Ω, and C = 2.4305 x 10-2 μF.80.sI − A =s 00 s⎡⎣⎢⎤⎦⎥ −(3.45 −14000Kc) −0.255x10−90.499x1011−3.68⎡⎣⎢⎤⎦⎥=s − (3.45 −14000Kc) 0.255x10−9−0.499x1011s + 3.68⎡⎣⎢⎤⎦⎥sI − A = s2+ (0.23 + 0.14x105Kc )s + (51520Kc + 0.0285)(2ζωn )2= [2* 0.9]2*(51520Kc + 0.0285) = (0.23 + 0.14x105Kc)2orKc2− 8.187x10−4Kc − 2.0122x10−10= 0Solving for Kc,Kc = 8.189x10−481.a. The transfer function from Chapter 2 is,Yh(s) − Ycat (s)Fup(s)=0.7883(s + 53.85)(s2+ 15.47s + 9283)(s2+8.119s + 376.3)The dominant poles come from s2+ 8.119s + 376.3. Using this polynomial,2ζωn = 8.119, and ωn2= 376.3. Thus, ωn = 19.4 and ζ = 0.209. Using Eq. (4.38), %OS =51.05%. Also,Ts =4ζωn= 0.985 s, and Tp =πωn 1− ζ2= 0.166 s. To find rise time, useFigure 4.16. Thus,ωnTr = 1.2136 or Tr = 0.0626 s.b. The other poles have a real part of 15.47/2 = 7.735. Dominant poles have a real part of 8.119/2 =4.06. Thus, 7.735/4.06 = 1.91. This is not at least 5 times.c.Program:syms snumg=0.7883*(s+53.85);deng=(s^2+15.47*s+9283)*(s^2+8.119*s+376.3);G(s) transfer functionG=vpa(numg/deng,3);pretty(G)numg=sym2poly(numg);deng=sym2poly(deng);G=tf(numg,deng)step(G)
  • 183. 4-76 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.Computer response:ans =G(s) transfer function.788 s + 42.4------------------------------------------2 2(s + 15.5 s + 9280.) (s + 8.12 s + 376.)Transfer function:0.7883 s + 42.45----------------------------------------------------s^4 + 23.59 s^3 + 9785 s^2 + 8.119e004 s + 3.493e006The time response shows 58 percent overshoot, Ts = 0.86 s, Tp = 0.13 s, Tr = 0.05 s.82.a. In Problem 3.30 we had⎥⎦⎤⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−+−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡21*00000*0000*00000)(uukTvTvTvTTckTvTvdvTTβββμβββ&&&[ ]⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=vTTy *100When 02 =u the equations are equivalent to
  • 184. Solutions to Design Problems 4-77Copyright © 2011 by John Wiley & Sons, Inc.10000*0000*000)(uvTvTvTTckTvTvdvTT⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−+−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡βββμβββ&&&[ ]⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=vTTy *100Substituting parameter values one gets1**02.52.54.210000058.024.00217.00058.0004167.0uvTTvTT⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡&&&[ ]⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=vTTy *100b.)det()(4.210000058.024.00217.00058.0004167.0)(11AIAIAI−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+−−+−+=−−−ssAdjssss[ ])0048.00398.0)(6419.2(0126.011.06817.2)17.2)(0058.0(58.0)4.2)(24.0()04167.0(100024.00217.00058.04.21000058.024.0)04167.0()det(223+++=+++=+−+++=−+−++−−++=−ssssssssssssss AITo obtain the adjoint matrix we calculate the cofactors:)64.2(4.21000058.024.011 +=+−−+= ssssC)4.2(0217.04.200058.00217.012 +−=+−−= ssC17.2100024.00217.013 =−+−=sC58.04.21000058.0021 =+−=sC
  • 185. 4-78 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.)4.2)(04167.0(4.200058.004167.022 ++=++= ssssC)04167.0(1001000004167.023 +−=−+= ssC)24.0(0058.00058.024.00058.0031 +−=−+= ssC1001.04117.24.20217.00058.004167.0 232 ++=+−+= ssssC01.02817.024.00217.0004167.0 233 ++=+−+= ssssCThen we have⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+++++−++++−−+=−01.02817.0)04167.0(10017.2)1101.04417.2()4.2)(04167.0()4.2(0217.0)24.0(0058.058.0)64.2()(22ssssssssssssAdj AIFinally[ ])0048.00398.0)(6419.2(02.0520)0048.00398.0)(6419.2(3844.1052002.52.5)0048.00398.0)(6419.2(01.028171.0)04167.0(10017.2)()(222211++++−=+++−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−++++++=−= −ssssssssssssssssUYBAICc. 100% effectiveness means that 11 =u orssU1)(1 = , so by the final value theorem820.11681)0048.00398.0)(6419.2()02.0(520)()( 200−=++++−==∞→→ ssssssLimssYLimyss(virus copies per mL of plasma)The closest poles to the imaginary axis are 0661.00199.0 j±− so the approximate settling timewill be 2100199.04=≈sT days.83.a.Substitutings2650F(s) =Δ into the transfer function and solving for ΔV(s) gives:
  • 186. Solutions to Design Problems 4-79Copyright © 2011 by John Wiley & Sons, Inc.)101908()101908(26501908F(s)V(s)+⋅+=+⋅=⋅Δ=ΔsBsAsssHere: 2650)101908(2650==+⋅=ssA and2650505,6201190.8Bs s= = −= −Substituting we have:⎟⎟⎠⎞⎜⎜⎝⎛×+−=+⋅−=Δ −)1024.5(11265)101908(505620265V(s) 3ssssTaking the inverse Laplace transform, we have:m/sin),()1(265)(31024.5tutetv ⋅−=Δ−×−b.>> s=tf(s);>> G=1/(1908*s+10);>> t=0:0.1:1000;>> y1=2650*step(G,t);>> y2=265*(1-exp(-5.24e-3.*t));>> plot(t,y1,t,y2)>> xlabel(sec)>> ylabel(m/s)
  • 187. 4-80 Chapter 4: Time ResponseCopyright © 2011 by John Wiley & Sons, Inc.0 100 200 300 400 500 600 700 800 900 1000050100150200250300secm/sBoth plots are identical.
  • 188. Copyright © 2011 by John Wiley & Sons, Inc.F I V EReduction of MultipleSubsystemsSOLUTIONS TO CASE STUDIES CHALLENGESAntenna Control: Designing a Closed-Loop Responsea. Drawing the block diagram of the system:+-10ΠiuK 150s+150uo0.16s(s+1.32)Pots Pre ampPowerampMotor,load andgearsThus, T(s) =76.39Ks3+151.32s2+198s+76.39Kb. Drawing the signal flow-diagram for each subsystem and then interconnecting them yields:10ΠiuKu o1s1s-150 -1.32150 0.81s 0.210Π-potpotpreamppowerampmotorandloadgearsx12x3x
  • 189. 5-2 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.x.1 = x2x.2 = - 1.32x2 + 0.8x3x.3 = -150x3 +150K(10π(qi − 0.2x1)) = -95.49Kx1 - 150x3 + 477.46Kθiθo = 0.2x1In vector-matrix notation,x.=0 1 00 -1.32 0.8-95.49K 0 -150x +00477.46Kθiθo = 0.2 0 0 xc. T1 =10π⎛⎝⎞⎠(K)(150)1s⎛⎝⎞⎠(0.8)1s⎛⎝⎞⎠1s⎛⎝⎞⎠(0.2) =76.39s3GL1 =−150s; GL 2 =−1.32s; GL3 = (K)(150)1s⎛⎝⎞⎠(0.8)1s⎛⎝⎞⎠1s⎛⎝⎞⎠(0.2)−10π⎛⎝⎞⎠=−76.39Ks3Nontouching loops:GL1GL2 =198s2Δ = 1 - [GL1 + GL2 + GL3] + [GL1GL2] = 1 +150s +1.32s +76.39Ks3 +198s2Δ1 = 1T(s) =T1Δ1Δ=76.39Ks3+151.32s2 +198s+76.39Kd. The equivalent forward path transfer function is G(s) =10π0.16Ks(s+1.32) .Therefore,T(s) =2.55s2+1.32s+2.55The poles are located at -0.66 ± j1.454. ωn = 2.55 = 1.597 rad/s; 2ζωn = 1.32, therefore, ζ = 0.413.
  • 190. Solutions to Case Studies Challenges 5-3Copyright © 2011 by John Wiley & Sons, Inc.%OS = e−ζπ / 1−ζ2x100 = 24%; Ts =4ζωn=40.66 = 6.06 seconds; Tp =πωn 1-ζ2=π1.454 =2.16 seconds; Using Figure 4.16, the normalized rise time is 1.486. Dividing by the natural frequency,Tr =1.4862.55= 0.93 seconds.e.f. Since G(s) =0.51Ks(s+1.32) , T(s) =0.51Ks2+1.32s+0.51K. Also, ζ =- ln (%OS100 )π2 + ln2 (%OS100 )= 0.517 for 15%overshoot; ωn = 0.51K ; and 2ζωn = 1.32. Therefore, ωn =1.322ζ=1.322(0.5147) = 1.277 = 0.51K .Solving for K, K=3.2.UFSS Vehicle: Pitch-Angle Control Representationa. Use the observer canonical form for the vehicle dynamics so that the output yaw rate is a statevariable.-22-1 -0.1251s 11s11-1u1s 1yx10.4371s1-0.24897-1.483x2x3x4
  • 191. 5-4 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.b. Using the signal flow graph to write the state equations:1 22 2 3 43 2 44 1 2 41.483 0.1250.24897 (0.125*0.437)2 2 2 2x xx x x xx x xx x x x u== − + −= − −= + − −&&&&In vector-matrix form:0 1 0 0 00 1.483 0 0.125 00 0.24897 0 0.054625 02 2 0 2 2[1 0 0 0]y⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥+⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦=x xx& =c.Program:numg1=-0.25*[1 0.437];deng1=poly([-2 -1.29 -0.193 0]);G(s)G=tf(numg1,deng1)numh1=[-1 0];denh1=[0 1];H(s)H=tf(numh1,denh1)Ge(s)Ge=feedback(G,H)T(s)T=feedback(-1*Ge,1)[numt,dent]=tfdata(T,V);[Acc,Bcc,Ccc,Dcc]=tf2ss(numt,dent)Computer response:ans =G(s)Transfer function:-0.25 s - 0.1093--------------------------------------s^4 + 3.483 s^3 + 3.215 s^2 + 0.4979 sans =H(s)Transfer function:-sans =Ge(s)
  • 192. Answers to Review Questions 5-5Copyright © 2011 by John Wiley & Sons, Inc.Transfer function:-0.25 s - 0.1093--------------------------------------s^4 + 3.483 s^3 + 3.465 s^2 + 0.6072 sans =T(s)Transfer function:0.25 s + 0.1093-----------------------------------------------s^4 + 3.483 s^3 + 3.465 s^2 + 0.8572 s + 0.1093Acc =-3.4830 -3.4650 -0.8572 -0.10931.0000 0 0 00 1.0000 0 00 0 1.0000 0Bcc =1000Ccc =0 0 0.2500 0.1093Dcc =0ANSWERS TO REVIEW QUESTIONS1. Signals, systems, summing junctions, pickoff points2. Cascade, parallel, feedback3. Product of individual transfer functions, sum of individual transfer functions, forward gain divided byone plus the product of the forward gain times the feedback gain4. Equivalent forms for moving blocks across summing junctions and pickoff points5. As K is varied from 0 to ∞, the system goes from overdamped to critically damped to underdamped.When the system is underdamped, the settling time remains constant.6. Since the real part remains constant and the imaginary part increases, the radial distance from the originis increasing. Thus the angle θ is increasing. Since ζ= cos θ the damping ratio is decreasing.7. Nodes (signals), branches (systems)8. Signals flowing into a node are added together. Signals flowing out of a node are the sum of signalsflowing into a node.9. One10. Phase-variable form, cascaded form, parallel form, Jordan canonical form, observer canonical form
  • 193. 5-6 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.11. The Jordan canonical form and the parallel form result from a partial fraction expansion.12. Parallel form13. The system poles, or eigenvalues14. The system poles including all repetitions of the repeated roots15. Solution of the state variables are achieved through decoupled equations. i.e. the equations are solvableindividually and not simultaneously.16. State variables can be identified with physical parameters; ease of solution of some representations17. Systems with zeros18. State-vector transformations are the transformation of the state vector from one basis system to another.i.e. the same vector represented in another basis.19. A vector which under a matrix transformation is collinear with the original. In other words, the lengthof the vector has changed, but not its angle.20. An eigenvalue is that multiple of the original vector that is the transformed vector.21. Resulting system matrix is diagonal.SOLUTIONS TO PROBLEMS1.a. Combine the inner feedback and the parallel pair.Multiply the blocks in the forward path and apply the feedback formula to get,T(s) =50(s-2)s3+s2+150s-100.b.Program:G1(s)G1=tf(1,[1 0 0])G2(s)G2=tf(50,[1 1])G3(s)G3=tf(2,[1 0])G4(s)G4=tf([1 0],1)G5(s)G5=2Ge1(s)=G2(s)/(1+G2(s)G3(s))Ge1=G2/(1+G2*G3)Ge2(s)=G4(s)-G5(s)
  • 194. Solutions to Problems 5-7Copyright © 2011 by John Wiley & Sons, Inc.Ge2=G4-G5Ge3(s)=G1(s)Ge1(s)Ge2(s)Ge3=G1*Ge1*Ge2T(s)=Ge3(s)/(1+Ge3(s))T=feedback(Ge3,1);T=minreal(T)Computer response:ans =G1(s)Transfer function:1---s^2ans =G2(s)Transfer function:50-----s + 1ans =G3(s)Transfer function:2-sans =G4(s)Transfer function:sans =G5(s)G5 =2ans =Ge1(s)=G2(s)/(1+G2(s)G3(s))Transfer function:50 s^2 + 50 s-------------------------s^3 + 2 s^2 + 101 s + 100ans =Ge2(s)=G4(s)-G5(s)
  • 195. 5-8 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Transfer function:s - 2ans =Ge3(s)=G1(s)Ge1(s)Ge2(s)Transfer function:50 s^3 - 50 s^2 - 100 s-------------------------------s^5 + 2 s^4 + 101 s^3 + 100 s^2ans =T(s)=Ge3(s)/(1+Ge3(s))Transfer function:50 s - 100-----------------------s^3 + s^2 + 150 s - 1002.Push G1(s) to the left past the pickoff point.+-++G1H1G11G2 G3Thus, T(s) =G11 + G1 H1⎛⎝⎜⎜⎞⎠⎟⎟ G2 +1G1⎛⎝⎜⎜⎞⎠⎟⎟G3 =G1G2 +1( )G31+ G1H1( )
  • 196. Solutions to Problems 5-9Copyright © 2011 by John Wiley & Sons, Inc.3.a. Split G3 and combine with G2 and G4. Also use feedback formula on G6 loop.Push G2 +G3 to the left past the pickoff point.Using the feedback formula and combining parallel blocks,Multiplying the blocks of the forward path and applying the feedback formula,
  • 197. 5-10 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.4.Push G2(s) to the left past the summing junction.Collapse the summing junctions and add the parallel transfer functions.Push G1(s)G2(s) + G5(s) to the right past the summing junction.
  • 198. Solutions to Problems 5-11Copyright © 2011 by John Wiley & Sons, Inc.Collapse summing junctions and add feedback paths.Applying the feedback formula,3 1 22 43 1 23 1 23 1 23 1 2 2 4( ) ( ) ( )( )( ) ( )1 [ ( ) ( ) ( )]( ) ( ) ( )( ) ( ) ( )1 [ ( ) ( ) ( )] ( ) ( )G s G s G sT sG s G sG s G s G s HG s G s G sG s G s G sH G s G s G s G s G s+=⎡ ⎤+ + +⎢ ⎥+⎣ ⎦+=+ + +
  • 199. 5-12 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.5.a. Push G7 to the left past the pickoff point. Add the parallel blocks, G3+G4.Push G3+G4 to the right past the summing junction.Collapse the minor loop feedback.
  • 200. Solutions to Problems 5-13Copyright © 2011 by John Wiley & Sons, Inc.PushG7(G3+G4)1+G6G7to the left past the pickoff point.Push G1 to the right past the summing junction.
  • 201. 5-14 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Add the parallel feedback paths to get the single negative feedback,H(s) =G5G7+G2(1+G6G7)G7(G3+G4) -G8G1. Thus,T(s) =b.Program:G1=tf([0 1],[1 7]); %G1=1/s+7 input transducerG2=tf([0 0 1],[1 2 3]); %G2=1/s^2+2s+3G3=tf([0 1],[1 4]); %G3=1/s+4G4=tf([0 1],[1 0]); %G4=1/sG5=tf([0 5],[1 7]); %G5=5/s+7G6=tf([0 0 1],[1 5 10]); %G6=1/s^2+5s+10G7=tf([0 3],[1 2]); %G7=3/s+2G8=tf([0 1],[1 6]); %G8=1/s+6G9=tf([1],[1]); %Add G9=1 transducer at the inputT1=append(G1,G2,G3,G4,G5,G6,G7,G8,G9);Q=[1 -2 -5 92 1 8 03 1 8 04 1 8 05 3 4 -66 7 0 07 3 4 -68 7 0 0];inputs=9;outputs=7;Ts=connect(T1,Q,inputs,outputs);T=tf(Ts)
  • 202. Solutions to Problems 5-15Copyright © 2011 by John Wiley & Sons, Inc.Computer response:Transfer function:6 s^7 + 132 s^6 + 1176 s^5 + 5640 s^4 + 1.624e004 s^3+ 2.857e004 s^2 + 2.988e004 s + 1.512e004-----------------------------------------------------------s^10 + 33 s^9 + 466 s^8 + 3720 s^7 + 1.867e004 s^6+ 6.182e004 s^5 + 1.369e005 s^4 + 1.981e005 s^3+ 1.729e005 s^2 + 6.737e004 s - 1.044e0046.Combine G6 and G7 yielding G6G7. Add G4 and obtain the following diagram:Next combine G3 and G4+G6G7.Push G5 to the left past the pickoff point.
  • 203. 5-16 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Notice that the feedback is in parallel form. Thus the equivalent feedback, Heq(s) =G2G5+G3(G4+G6G7) + G8. Since the forward path transfer function is G(s) = Geq(s) = G1G5, the closed-loop transfer function isT(s) =Geq(s)1+Geq(s)Heq(s) .Hence,7.Push 2s to the right past the pickoff point.
  • 204. Solutions to Problems 5-17Copyright © 2011 by John Wiley & Sons, Inc.Combine summing junctions.Combine parallel 2s and s. Apply feedback formula to unity feedback with G(s) = s.Combine cascade pair and add feedback around 1/(s+1).
  • 205. 5-18 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Combine parallel pair and feedback in forward path.Combine cascade pair and apply final feedback formula yielding T(s) =5s2+ 2s6s2+ 9s + 6.8.Push G3 to the left past the pickoff point. Push G6 to the left past the pickoff point.
  • 206. Solutions to Problems 5-19Copyright © 2011 by John Wiley & Sons, Inc.Hence,Thus the transfer function is the product of the functions, orθ22(s)θ11(s)=G1G2G4G5G6G71 - G4G5 + G4G5G6 + G1G2G3 - G1G2G3G4G5 + G1G2G3G4G5G69.Combine the feedback with G6 and combine the parallel G2 and G3.Move G2+G3 to the left past the pickoff point.
  • 207. 5-20 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Combine feedback and parallel pair in the forward path yielding an equivalent forward-path transferfunction ofGe(s) =⎝⎛⎠⎞G2+G31+G1(G2+G3) ⎝⎛⎠⎞G5+G4G2+G3 ⎝⎛⎠⎞G61+G6But, T(s) =Ge(s)1+Ge(s)G7(s) . Thus,10.Push G3(s) to the left past the pickoff point.
  • 208. Solutions to Problems 5-21Copyright © 2011 by John Wiley & Sons, Inc.Push G2(s)G3(s) to the left past the pickoff point.Push G1(s) to the right past the summing junction.
  • 209. 5-22 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Collapsing the summing junctions and adding the feedback transfer functions,T(s) =G1(s)G2(s)G3(s)1 + G1(s)G2 (s)G3(s)Heq (s)whereHeq (s) =H3 (s)G3(s)+H1(s)G2 (s)G3(s)+H2(s)G1(s)G3(s)+H4 (s)G1(s)+111.2225( )15 225T ss s=+ +. Therefore, 2ζωn = 15, and ωn = 15. Hence, ζ = 0.5.%OS = e−ζπ / 1−ζ2x100 = 16.3%; Ts =4ζωn=0.533; Tp =πωn 1-ζ2=0.242.12.2 22 25( )( 2 5) 2 515 2 51, 2A Bs CC ss s s s s sAs s Bs CsB C+= = ++ + + +== + + + +∴ = − = −2 22 21 2 1 2( )2 5 ( 1) 41( 1) 21 ( 1) 1 1 2( 1) 4 ( 1) 4s sC ss s s s ssss s s s+ += − = −+ + + ++ ++ += − = −+ + + +1( ) 1 (cos2 sin 2 )2tc t e t t−= − +
  • 210. Solutions to Problems 5-23Copyright © 2011 by John Wiley & Sons, Inc.13.Push 2s to the left past the pickoff point and combine the parallel combination of 2 and 1/s.Push (2s+1)/s to the right past the summing junction and combine summing junctions.Hence, T(s) =2(2s +1)1 + 2(2s +1)Heq (s), where Heq (s) = 1 +s2s + 1+52s.14.SinceG(s) =Ks(s + 30), T(s) =G(s)1 + G(s)=Ks2+ 30s + K. Therefore, 2ζωn = 30. Thus, ζ =15/ωn = 0.5912 (i.e. 10% overshoot). Hence, ωn = 25.37 = K . Therefore K = 643.6.15.T(s) =Ks2+ αs + K; ζ =−ln(%OS100)π2+ ln2(%OS100)= 0.358 ;40.15snTζω= = . Therefore, ωn =74.49. K = ωn2 = 5549. α = 2ζωn = 53.33.
  • 211. 5-24 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.16.The equivalent forward-path transfer function is G(s) =10K1s[s + (10K2 + 2)]. Hence,T(s) =G(s)1 + G(s)=10K1s2+ (10K2 + 2)s + 10K1. Sincep43.2, Re 1.25; and T 1.5, Im=2.09Re ImsTπ= = ∴ = = = ∴ . The poles are thus at–1.25+j2.09. Hence,2 211.25 2.09 10n Kω = + = . Thus, K1 = 0.593. Also, (10K2 + 2)/2 = Re= 1.25. Hence, K2 = 0.05.17.a. For the inner loop, Ge(s) =20s(s +12), and He(s) = 0.2s. Therefore, Te(s) =Ge(s)1 + Ge(s)He(s) =20s(s+16) . Combining with the equivalent transfer function of the parallel pair, Gp(s) = 20, the systemis reduced to an equivalent unity feedback system with G(s) = Gp(s) Te(s) =400s(s+16) . Hence, T(s) =G(s)1+G(s) =400s2+16s+400.b. ωn2 = 400; 2ζωn = 16. Therefore, ωn = 20, and ζ = 0.4. %OS = e−ζπ / 1−ζ2x100 = 25.38;Ts =4ζωn=0.5; Tp =πωn 1-ζ2=0.171. From Figure 4.16, ωnTr = 1.463. Hence, Tr = 0.0732.ωd = Im = ωn 1 - ζ2 = 18.33.18.238343( )200 38343T ss s=+ +; from which, 2ζωn = 200 and ωn = 38343 = 195.81. Hence,2- / 1-0.511. %OS=e 100 15.44%xζπ ζζ = = ;20.01871pnTπω ζ= =−s.Also,Ts =4ζωn= 0.04 s.19.For the generator, Eg(s) = KfIf(s). But, If(s) =Ei(s)Rf+Lfs . Therefore,Eg(s)Ei(s) =2s+1 . For the motor,consider Ra = 2 Ω, the sum of both resistors. Also, Je = Ja+JL(12 )2 = 0.75+4x14 = 1.75; De = DL(12 )2= 1. Therefore,
  • 212. Solutions to Problems 5-25Copyright © 2011 by John Wiley & Sons, Inc.θm(s)Eg(s) =1( ( ))ta et bee aKR JK Ks s DJ R+ +=0.286( 0.857)s s +.But,θo(s)θm(s)=12 . Thus,θo (s)Eg(s) =0.143( 0.857)s s +. Finally,θo(s)Ei(s) =Eg(s)Ei(s)θo (s)Eg(s) =0.286( 1)( 0.857)s s s+ +.20.For the mechanical system, J(N2N1)2s2θ2(s) = T(N2N1) . For the potentiometer, Ei (s)= 10θ2(s)2π, orθ2(s) =π5 Ei(s). For the network, Eo(s) = Ei(s)RR+1Cs= Ei(s)ss+1RC, or Ei(s) = Eo(s)s+1RCs .Therefore, θ2 (s) =π5Eo(s)s +1RCs. Substitute into mechanical equation and obtain,Eo (s)T(s)=5N1JπN2s s +1RC⎛⎝⎞⎠.21.The equivalent mechanical system is found by reflecting mechanical impedances to the spring.Writing the equations of motion:4s2+ 2s + 5( )θ1(s) − 5θ2(s) = 4T(s)−5θ1(s) + 2s2+ 5( )θ2(s) = 0Solving for θ2(s),
  • 213. 5-26 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.θ2 (s) =4s2+ 2s + 5( ) 4T(s)−5 04s2+ 2s + 5( ) −5−5 2s2+ 5( )=20T(s)8s4+ 4s3+ 30s2+10sThe angular rotation of the pot is 0.2 that of θ2, orθp(s)T(s)=2s 4s3+ 2s2+15s + 5( )For the pot:Ep (s)θp(s)=505(2π )=5πFor the electrical network: Using voltage division,Eo (s)Ep (s)=200,000110−5s+ 200,000=ss +12Substituting the previously obtained values,Eo(s)T(s)=θp (s)T(s)⎛⎝⎜⎞⎠⎟Ep (s)θp (s)⎛⎝⎜⎜⎞⎠⎟⎟Eo (s)Ep(s)⎛⎝⎜⎜⎞⎠⎟⎟ =10πss s +12⎛⎝⎜⎞⎠⎟ 4s3+ 2s2+15s + 5( )22.a.
  • 214. Solutions to Problems 5-27Copyright © 2011 by John Wiley & Sons, Inc.r x1x2x3x411s250s +1 s2−2s−1b.x1x2x3x4x5x5x4x3x2x1
  • 215. 5-28 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.r x5x4x 3x2 x11G3G4G611G15GG7---1G2c.x5x4x1x3 x2G65GG2G3G4G1 G7x1x2x 3x 4x 5r 118G---
  • 216. Solutions to Problems 5-29Copyright © 2011 by John Wiley & Sons, Inc.23.a.x1•= x2x2•= x3x3•= −2x1 − 4x2 − 6x3 + ry = x1 + x2-6-4-21111s1s1s x1x3x2ryb.x1•= x2x2•= −3x2 + x3 + rx3•= −3x1 − 4x2 − 5x3 + ry = x1 + 2x21
  • 217. 5-30 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.c.x1•= 7x1 + x2 + rx2•= −3x1 + 2x2 − x3 + 2rx3•= −x1 + 2x3 + ry = x1 + 3x2 + 2x3-3224.a. Since G(s) =10s3+ 24s2+191s + 504=C(s)R(s) ,c•••+ 24 c••+ 191c•+ 504c = 10rLet,c = x1c•= x2c••= x3Therefore,x1•= x2x2•= x3x3•= −504x1 −191x2 − 24x3 +10ry = x1
  • 218. Solutions to Problems 5-31Copyright © 2011 by John Wiley & Sons, Inc.1s1s1s10 1-24-191r x1x2x3 y-504b. G(s) = (10s + 7) (1s +8) (1s + 9)1s1s1s10 1r x1x2x3 y1 1-8 -9-7Therefore,x1•= −9x1 + x2x2•= −8x2 + x3x3•= −7x3 +10ry = x125.a. Since G(s) =20s4+15s3+ 66s2+ 80s=C(s)R(s) ,c••••+15 c•••+ 66 c••+80 c•= 20rLet,c = x1c•= x2c••= x3c•••= x4Therefore,
  • 219. 5-32 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.x1•= x2x2•= x3x3•= x4x4•= −80x2 − 66x3 −15x4 + 20ry = x11s1s1s20 1-15-66rx1x2x3 y-801sx4b. G(s) = (20s) (1s + 2) (1s + 5) (1s +8). Hence,1s1s1s20 1r x1x2x3 y1 1-2 -5x4-811sFrom which,x1•= −8x1 + x2x2•= −5x2 + x3x3•= −2x3 + x4x4•= 20ry = x126.Δ = 1 + [G2G3G4 + G3G4 + G4 + 1] + [G3G4 + G4]; T1 = G1G2G3G4; Δ1 = 1. Therefore,T(s) =T1Δ1Δ=G1G2G3G42 + G2G3G4 + 2G3G4 + 2G4
  • 220. Solutions to Problems 5-33Copyright © 2011 by John Wiley & Sons, Inc.27.Closed-loop gains: G2G4G6G7H3; G2G5G6G7H3; G3G4G6G7H3; G3G5G6G7H3; G6H1; G7H2Forward-path gains: T1 = G1G2G4G6G7; T2 = G1G2G5G6G7; T3 = G1G3G4G6G7; T4 =G1G3G5G6G7Nontouching loops 2 at a time: G6H1G7H2Δ = 1 - [H3G6G7(G2G4 + G2G5 + G3G4 + G3G5) + G6H1 + G7H2] + [G6H1G7H2]Δ1 = Δ2 = Δ3 = Δ4 = 1T(s) =T1Δ1 + T2Δ2 + T3Δ3 + T4Δ4Δ=G1G2G4G6G7 + G1G2G5G6G7 + G1G3G4G6G7 + G1G3G5G6G71 - H3G6G7(G2G4 + G2G5 + G3G4 + G3G5) - G6H1 - G7H2 + G6H1G7H228.Closed-loop gains: -s2; -1s ; -1s ; -s2Forward-path gains: T1 = s; T2 =1s2Nontouching loops: NoneΔ = 1 - (-s2 -1s -1s - s2)Δ1 = Δ2 = 1G(s) =T1Δ1 + T2Δ2Δ=s +1s21 + (s2 +1s +1s + s2)=s3+12s4+s2+2s29.T(s) =G1⎝⎛⎠⎞G2G3G4G5(1-G2H1)(1-G4H2)1 -G2G3G4G5G6G7G8(1-G2H1)(1-G4H2)(1-G7H4)=G1G2G3G4G5(1-G7H4)1-G2H1-G4H2+G2G4H1H2-G7H4+G2G7H1H4+G4G7H2H4-G2G4G7H1H2H4-G2G3G4G5G6G7G830.a. G(s) =(s +1)(s + 2)(s + 3)2(s + 4)=2(s + 3)2 −5s + 3+6s + 4
  • 221. 5-34 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Writing the state and output equations,x.1 = -3x1 + x2x.2 = -3x2 + rx.3 = -4x3 + ry = 2x1 - 5x2 + 6x3In vector-matrix form,x•=−3 1 00 −3 00 0 −4⎡⎣⎢⎢⎤⎦⎥⎥x +011⎡⎣⎢⎢⎤⎦⎥⎥ry = 2 −5 6[ ]b. G(s) = G(s) =(s + 2)(s + 5)2(s + 7)2 = −3/4(s + 5)2 +1s + 5−5 /4(s + 7)2 −1s + 71s1s1s1s111-5 -5- 34- 54-1-7 -711rx2 x1x3x4yWriting the state and output equations,x.1 = -5x1 + x2
  • 222. Solutions to Problems 5-35Copyright © 2011 by John Wiley & Sons, Inc.x.2 = -5x2 + rx.3 = -7x3 + x4x.4 = -7x4 + ry = -34 x1 + x2 -54 x3 - x4In vector matrix form,x.=- 5 1 0 00 - 5 0 00 0 - 7 10 0 0 - 7x +0101ry = - 341 - 54- 1 xc.Writing the state and output equations,x.1 = - 2x1 + x2x.2 = - 2x2 + rx.3 = - 4x3 + rx.4 = - 5x4 + ry =16 x1 +136 x2 -14 x3 +29 x4In vector-matrix form,
  • 223. 5-36 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.31.a.Writing the state equations,x.1 = x2x.2 = - 7x1 - 2x2 + ry = 3x1 + x2In vector matrix form,
  • 224. Solutions to Problems 5-37Copyright © 2011 by John Wiley & Sons, Inc.b.Writing the state equations,x1•= x2x2•= x3x3•= −x1 − 2x2 − 5x3 + ry = 6x1 + 2x2 + x3In vector matrix form,x•=0 1 00 0 1−1 −2 −5⎡⎣⎢⎢⎤⎦⎥⎥X +001⎡⎣⎢⎢⎤⎦⎥⎥ry = 6 2 1[ ]x
  • 225. 5-38 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.c.x.1 = x2x.2 = x3x.3 = x4x.4 = - 4x1 - 6x2 - 5x3 - 3x4 + ry = x1 + 7x2 + 2x3 + x4In vector matrix form,32.a. Controller canonical form:From the phase-variable form in Problem 5.31(a), reverse the order of the state variables and obtain,
  • 226. Solutions to Problems 5-39Copyright © 2011 by John Wiley & Sons, Inc.x.2 = x1x.1 = - 7x2 - 2x1 + ry = 3x2 + x1Putting the equations in order,x.1 = - 2x1 - 7x2 + rx.2 = x1y = x1 + 3x2In vector-matrix form,x•=−2 −71 0⎡⎣⎢⎤⎦⎥x +10⎡⎣⎢⎤⎦⎥ry = 1 3[ ]xObserver canonical form:G(s) =s+3s2+2s+7. Divide each term by1s2 and getG(s) =1s+3s21 +2s+7s2=C(s)R(s)Cross multiplying,(1s +3s2 ) R(s) = (1 +2s +7s2 ) C(s)Thus,1s (R(s) - 2C(s)) +1s2 (3R(s) - 7C(s)) = C(s)Drawing the signal-flow graph,
  • 227. 5-40 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.-7-231R(s)Writing the state and output equations,x.1 = - 2x1 + x2 + rx.2 = - 7x1 + 3ry = x1In vector matrix form,x•=−2 1−7 0⎡⎣⎢⎤⎦⎥x +13⎡⎣⎢⎤⎦⎥ry = 1 0[ ]xb. Controller canonical form:From the phase-variable form in Problem 5.31(b), reverse the order of the state variables and obtain,x3•= x2x2•= x1x1•= −x3 − 2x2 − 5x1y = 6x3 + 2x2 + x1Putting the equations in order,x1•= −5x1 − 2x2 − x3x2•= x1x3•= x2y = x1 + 2x2 + 6x3In vector-matrix form,
  • 228. Solutions to Problems 5-41Copyright © 2011 by John Wiley & Sons, Inc.x•=−5 −2 −11 0 00 1 0⎡⎣⎢⎢⎤⎦⎥⎥x +100⎡⎣⎢⎢⎤⎦⎥⎥ry = 1 2 6[ ]xObserver canonical form:G(s) =s2+ 2s + 6s3+ 5s2+ 2s +1. Divide each term by1s3 and getG(s) =1s+2s2+6s31 +5s+2s2+1s3=C(s)R(s)Cross-multiplying,1s+2s2 +6s3⎛⎝⎞⎠R(s) = 1 +5s+2s2 +1s3⎛⎝⎞⎠C(s)Thus,1s(R(s)− 5c(s)) +1s2 (2R(s) − 2C(s)) +1s3 (6R(s) − C(s)) = C(s)Drawing the signal-flow graph,1s1s1s6 1 1 1R(s) C(s)X1(s)X2(s)X3(s)21-5-2-1Writing the state and output equations,x1•= −5x1 + x2 + rx2•= −2x1 + x3 + 2rx3•= −x1 + 6ry = 1 0 0[ ]x
  • 229. 5-42 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.In vector-matrix form,x•=−5 1 0−2 0 1−1 0 0⎡⎣⎢⎢⎤⎦⎥⎥x +126⎡⎣⎢⎢⎤⎦⎥⎥ry = 1 0 0[ ]xc. Controller canonical form:From the phase-variable form in Problem 5.31(c), reverse the order of the state variables and obtain,x.4 = x3x.3 = x2x.2 = x1x.1 = - 4x4 - 6x3 - 5x2 - 3x1 + ry = x4 + 7x3 + 2x2 + x1Putting the equations in order,x.1 = - 3x1 - 5x2 - 6x3 - 4x4 + rx.2 = x1x.3 = x2x.4 = x3y = x1 + 2x2 +7x3 + x4In vector-matrix form,x•=−3 −5 −6 −41 0 0 00 1 0 00 0 1 0⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥X +1000⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥ry = 1 2 7 1[ ]xObserver canonical form:G(s) =s3+2s2+7s+1s4+3s3+5s2+6s+4. Divide each term by1s2 and getG( s ) =1s+2s2+7s3+1s41 +3s+5s2+6s3+4s4=C( s )R( s )
  • 230. Solutions to Problems 5-43Copyright © 2011 by John Wiley & Sons, Inc.Cross multiplying,(1s +2s2 +7s3 +1s4 ) R(s) = (1 +3s +5s2 +6s3 +4s4 ) C(s)Thus,1s (R(s) - 3C(s)) +1s2 (2R(s) - 5C(s)) +1s3 (7R(s) - 6C(s)) +1s4 (R(s) - 4C(s)) = C(s)Drawing the signal-flow graph,R(s)-4-6-5-31271Writing the state and output equations,x.1 = - 3x1 + x2 + rx.2 = - 5x1 + x3 + 2rx.3 = - 6x1 + x4 +7rx.4 = - 4x1 + ry = x1In vector matrix form,
  • 231. 5-44 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.33.a.1s1s1s1 50 1 1-5 -7-1r c= yx12x3x-2Writing the state equations,1 1 22 2 33 1 312850 9 50x x xx x xx x x ry x•••= − += − += − − +=In vector-matrix form,[ ]2 1 0 00 8 1 050 0 9 501 0 0ry•−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦=x xxb.x1c= y1s1s2x10-8-25-1r11s3x-2-8-9-6-24
  • 232. Solutions to Problems 5-45Copyright © 2011 by John Wiley & Sons, Inc.Writing the state equations,1 22 1 2 33 1124 6 10x xx x x xx x ry x•••== − − += − +=In vector-matrix form,[ ]0 1 0 024 6 10 01 0 0 11 0 0ry•⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − − +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦=x xxc.r c= yx12x1s1s1 100 1 1-1-1-1Tach feedbackbefore integratorx.1 = x2x.2 = -x2 - x2 + 160(r-x1) = -160x1 -2x2 +160ry = x1In vector-matrix form,0 1 0160 2 160r⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦xx&y = 1 0 x160
  • 233. 5-46 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.d. Since1(s+1)2 =1s2+2s+1, we draw the signal-flow as follows:r c= yx12x1s1s1 210-2-1-11Writing the state equations,x.1 = x2x.2 = -x1 - 2x2 + 16(r-c) = -x1 - 2x2 + 16(r - (2x1+x2) = -33x1 - 18x2 + 16ry = 2x1 + x2In vector-matrix form,0 1 0-33 -18 16r⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦xx&[2 1]y =& x34.a. Phase-variable form:T(s) =10s3+3s2+2s+10r= u c= yx12x1s1s10-2-3-101s3xWriting the state equations,x.1 = x2x.2 = x3x.3 = -10x1 -2x2 -3x3 + 10uy = x1In vector-matrix form,16
  • 234. Solutions to Problems 5-47Copyright © 2011 by John Wiley & Sons, Inc.x0 1 00 0 1-10 -2 -3x +0010u=y = 1 0 0 xb. Parallel form:G(s) =5s +-10s+1 +5s+22x3xx1r=u c=y5-105111-1-21s1s1s-11r=uWriting the state equations,x.1 = 5(u - x1 - x2 - x3) = -5x1 -5x2 -5x3 +5ux.2 = -10(u - x1 - x2 - x3) - x2 = 10x1 + 9x2 + 10x3 - 10ux.3 = 5(u - x1 - x2 - x3) - 2x3 = -5x1 -5x2 -7x3 +5uy = x1 + x2 + x3In vector-matrix form,x•=−5 −5 −510 9 10−5 −5 −7⎡⎣⎢⎢⎤⎦⎥⎥x +5−105⎡⎣⎢⎢⎤⎦⎥⎥uy = 1 1 1[ ]x35.a. T(s) =10(s2+ 5s + 6)s4+16s3+ 99s2+ 244s +180
  • 235. 5-48 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Drawing the signal-flow diagram,3x4x1syx 12x101s1s1s 6r151699244180Writing the state and output equations,x1•= x2x2•= x3x3•= x4x1•= −180x1 − 244x2 − 99x3 −16x4 + 10ry = 6x1 + 5x2 + x3In vector-matrix form,x•=0 1 0 00 0 1 00 0 0 1−180 −244 −99 −16⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥x +00010⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥ry = 6 5 1 0[ ]xb. G(s) =10(s + 2)(s + 3)(s +1)(s + 4)(s + 5)(s + 6)=1/ 3s +1−10 /3s + 4+15s + 5−12s + 6Drawing the signal-flow diagram and including the unity-feedback path,
  • 236. Solutions to Problems 5-49Copyright © 2011 by John Wiley & Sons, Inc.3x4xx 12x1s1− 103 1r = u y1s1s-1-4-5-6-1-121531 11111sWriting the state and output equations,x1•=13(u − x1 − x2 − x3 − x4 ) − x1x2•=−103(u − x1 − x2 − x3 − x4 ) − 4x2x3•=15(u − x1 − x2 − x3 − x4 ) − 5x3x4•= −12(u − x1 − x2 − x3 − x4 ) − 12x4y = x1 + x2 + x3 + x4In vector-matrix form,x•=−43−13−13−13103−23103103−15 −15 −20 −1512 12 12 0⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥x +13−10315−12⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥uy = 1 1 1 1[ ]x36.Program:(a)G(s)G=zpk([-2 -3],[-1 -4 -5 -6],10)
  • 237. 5-50 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.T(s)T=feedback(G,1,-1)[numt,dent]=tfdata(T,v);Find controller canonical form[Acc,Bcc,Ccc,Dcc]=tf2ss(numt,dent)A1=flipud(Acc);Transform to phase-variable formApv=fliplr(A1)Bpv=flipud(Bcc)Cpv=fliplr(Ccc)(b)G(s)G=zpk([-2 -3],[-1 -4 -5 -6],10)T(s)T=feedback(G,1,-1)[numt,dent]=tfdata(T,v);Find controller canonical form[Acc,Bcc,Ccc,Dcc]=tf2ss(numt,dent)Transform to modal form[A,B,C,D]=canon(Acc,Bcc,Ccc,Dcc,modal)Computer response:ans =(a)ans =G(s)Zero/pole/gain:10 (s+2) (s+3)-----------------------(s+1) (s+4) (s+5) (s+6)ans =T(s)Zero/pole/gain:10 (s+2) (s+3)------------------------------------------(s+1.264) (s+3.412) (s^2 + 11.32s + 41.73)ans =Find controller canonical formAcc =-16.0000 -99.0000 -244.0000 -180.00001.0000 0 0 00 1.0000 0 00 0 1.0000 0Bcc =1000
  • 238. Solutions to Problems 5-51Copyright © 2011 by John Wiley & Sons, Inc.Ccc =0 10.0000 50.0000 60.0000Dcc =0ans =Transform to phase-variable formApv =0 1.0000 0 00 0 1.0000 00 0 0 1.0000-180.0000 -244.0000 -99.0000 -16.0000Bpv =0001Cpv =60.0000 50.0000 10.0000 0ans =(b)ans =G(s)Zero/pole/gain:10 (s+2) (s+3)-----------------------(s+1) (s+4) (s+5) (s+6)ans =T(s)Zero/pole/gain:10 (s+2) (s+3)------------------------------------------(s+1.264) (s+3.412) (s^2 + 11.32s + 41.73)ans =Find controller canonical formAcc =-16.0000 -99.0000 -244.0000 -180.00001.0000 0 0 00 1.0000 0 00 0 1.0000 0
  • 239. 5-52 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Bcc =1000Ccc =0 10.0000 50.0000 60.0000Dcc =0ans =Transform to modal formA =-5.6618 3.1109 0 0-3.1109 -5.6618 0 00 0 -3.4124 00 0 0 -1.2639B =-4.11081.04681.31250.0487C =0.1827 0.6973 -0.1401 4.2067D =037.1s1s1s1s1 1-1-1-11r c= y4x3x2x1xWriting the state equations,
  • 240. Solutions to Problems 5-53Copyright © 2011 by John Wiley & Sons, Inc.x.1 = x2x.2 = - x1 + x3x.3 = x4x.4 = x1 - x2 + ry = -x1 + x2In vector-matrix form,x =0 1 0 0-1 0 1 00 0 0 11 -1 0 0x +0001ry = c = [-1 1 0 0] x38.a.θ..1 + 5θ.1 + 6θ1 - 3θ.2 - 4θ2 = 0-3θ.1 - 4θ1 + θ..2 + 5θ.2 + 5θ2 = Torθ..1= - 5θ.1 - 6θ1 + 3θ.2 + 4θ2θ..2 = 3θ.1 + 4θ1 - 5θ.2 - 5θ2 + TLetting, θ1 = x1 ; θ.1 = x2 ; θ2 = x3 ; θ.2 = x4 ,-51s1s-6-51s1s-51431x2xT343x4x
  • 241. 5-54 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.where x = θ.b. Using the signal-flow diagram,x.1 = x2x.2 = -6x1 - 5x2 + 4x3 + 3x4x.3 = x4x.4 = 4x1 + 3x2 - 5x3 - 5x4 + Ty = x3In vector-matrix form,x.=0 1 0 0-6 -5 4 30 0 0 14 3 -5 -5x +0001Ty = 0 0 1 0 x39.Program:numg=7;deng=poly([0 -9 -12]);G=tf(numg,deng);T=feedback(G,1)[numt,dent]=tfdata(T,v)[A,B,C,D]=tf2ss(numt,dent); %Obtain controller canonical form(a) %Display labelA=flipud(A); %Convert to phase-variable formA=fliplr(A) %Convert to phase-variable formB=flipud(B) %Convert to phase-variable formC=fliplr(C) %Convert to phase-variable form(b) %Display label[a,b,c,d]=canon(A,B,C,D) %Convert to parallel formComputer response:Transfer function:7------------------------s^3 + 21 s^2 + 108 s + 7numt =0 0 0 7dent =1 21 108 7
  • 242. Solutions to Problems 5-55Copyright © 2011 by John Wiley & Sons, Inc.ans =(a)A =0 1 00 0 1-7 -108 -21B =001C =7 0 0ans =(b)a =-0.0657 0 00 -12.1807 00 0 -8.7537b =-0.0095-3.58572.5906c =-6.9849 -0.0470 -0.0908d =040.x.1 = A1x1 + B1r (1)y1 = C1x1 (2)x.2 = A2x2 + B2y1 (3)y2 = C2x2 (4)Substituting Eq. (2) into Eq. (3),x.1 = A1x1 + B1r
  • 243. 5-56 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.x.2 = B2C1x1 + A2x2y2 = C2x2In vector-matrix notation,x1x2=A1B2 C1-OA2x1x 2+ B1Ory 2 = O C2x1x241.x.1 = A1x1 + B1r (1)y1 = C1x1 (2)x.2 = A2x2 + B2r (3)y2 = C2x2 (4)In vector-matrix form,=A1B2-OA2x 1x 2+ B1 rOx1x2y = y1 + y2 = C1 C2x1x 242.x.1 = A1x1 + B1e (1)y = C1x1 (2)x.2 = A2x2 + B2y (3)p = C2x2 (4)Substituting e = r - p into Eq. (1) and substituting Eq. (2) into (3), we obtain,x.1 = A1x1 + B1(r - p) (5)y = C1x1 (6)x.2 = A2x2 + B2C1x1 (7)p = C2x2 (8)Substituting Eq. (8) into Eq. (5),x.1 = A1x1 - B1C2x2 + B1r
  • 244. Solutions to Problems 5-57Copyright © 2011 by John Wiley & Sons, Inc.x.2 = B2C1x1 + A2x2y = C1x1In vector-matrix form,x1x2= A1B2 C1B1 C2A2x1x2+ B10r-x1x2y = C1 043.z•= P−1APz + P−1Buy = CPz14 9 3 -0.2085 -0.3029 -0.16610 4 7 ; = 0.0228 -0.1075 -0.09121 4 9 0.0130 0.0814 -0.0521−− −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − ∴⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦P P[ ]1 -118.5961 25.4756 5.6156 58-12.9023 -28.8893 -8.3909 ; 63 ; C = 1.5668 3.0423 2.7329-0.5733 11.4169 5.2932 12−−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦P AP P B P44.5 4 9 0.3469 -0.3878 0.26536 7 6 ; P= 0.3673 -0.4694 0.16336 5 3 0.0816 0.0068 -0.0748=-28.2857 40.8095 -40.9048-18.3061 28.2245 -37.46945.3878 -6.5510 -5.9388uy= +=−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎡ ⎤⎢= ⎢⎢⎣•-1 -1-1-1z P APz P BCPzPP AP [ ]61; 82 ; -2.0816 2.6599 -1.258574⎡ ⎤⎥ ⎢ ⎥= =⎥ ⎢ ⎥⎥ ⎢ ⎥⎦ ⎣ ⎦-1P B CP45.Eigenvalues are -1, -2, and -3 since,|λΙ - A | = (λ + 3) (λ + 2) (λ + 1)Solving for the eigenvectors, Ax = λxor,
  • 245. 5-58 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.For λ = -1, x2 = 0, x1 = x3 . For λ = -2, x1 = x2 =x32 . For λ = -3, x1 = -x22 , x2 = x3 . Thus,z.= P-1APz + P-1Bu ; y = CPz, where46.Eigenvalues are 1, -2, and 3 since,|λI - A | = (λ - 3) (λ + 2) (λ - 1)Solving for the eigenvectors, Ax = λxor,For λ = 1, x1 = x2 =x32 . For λ = -2, x1 = 2x3, x2 = -3x3. For λ = 3, x1 = x3 , x2 = -2x3 . Thus,z.= P-1APz + P-1Bu ; y = CPz, where47.Program:A=[-10 -3 7;18.25 6.25 -11.75;-7.25 -2.25 5.75];B=[1;3;2];C=[1 -2 4];[P,d]=eig(A);Ad=inv(P)*A*PBd=inv(P)*BCd=C*P
  • 246. Solutions to Problems 5-59Copyright © 2011 by John Wiley & Sons, Inc.Computer response:Ad =-2.0000 0.0000 0.0000-0.0000 3.0000 -0.00000.0000 0.0000 1.0000Bd =1.8708-3.67423.6742Cd =3.2071 3.6742 2.857748.a. Combine G1(s) and G2(s). Then push K1 to the right past the summing junction:1G (s)2G (s)1+-sK1C(s)s2+- -+R(s)K1K2Push K1K2 to the right past the summing junction:
  • 247. 5-60 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.1G (s)2G (s)1K1K2+ +-- -sK1+R(s) C(s)s2K1K2Hence, T(s) =K1K2G1(s)G2(s)1 + K1K2G1(s)G2(s)⎝⎜⎛⎠⎟⎞1 +sK1+s2K1K2b. Rearranging the block diagram to show commanded pitch rate as the input and actual pitch rate asthe output:Actualpitchrate1G (s)2G (s)1+-ss2-+CommandedpitchrateK2-K1Pushing K2 to the right past the summing junction; and pushing s to the left past the pick-off pointyields,
  • 248. Solutions to Problems 5-61Copyright © 2011 by John Wiley & Sons, Inc.1G (s)2G (s)1K2+ +--sK1sK2sActualpitchrateCommandedpitchrate-Finding the closed-loop transfer function:T(s) =K2sG1(s)G2(s)1 + K2sG1(s)G2(s)⎝⎛⎠⎞1 +sK2+K1s=K2sG1(s)G2(s)1 + G1(s)G2(s)(s2 + K2s + K1K2)c. Rearranging the block diagram to show commanded pitch acceleration as the input and actual pitchacceleration as the output:Actualpitchacceleration1G (s)2G (s)1+s2K2-+Commandedpitchacceleration-K1K2s-Pushing s2 to the left past the pick-off point yields,
  • 249. 5-62 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Actualpitchacceleration1G (s)2G (s)1+s2K2-+Commandedpitchacceleration-K1K2s-s2Finding the closed-loop transfer function:T(s) =s2G1(s)G2(s)1 + s2G1(s)G2(s)⎝⎜⎛⎠⎟⎞1 +K1K2s2 +K2s=s2G1(s)G2(s)1 + G1(s)G2(s)(s2 + K2s + K1K2)49.Establish a sinusoidal model for the carrier: T(s) =K1s2+a2x12x1s1s-ar 1 K1Establish a sinusoidal model for the message: T(s) =K2s2+b21s1s 3x4xr 2 K2-bWriting the state equations,
  • 250. Solutions to Problems 5-63Copyright © 2011 by John Wiley & Sons, Inc.x.1 = x2x.2 = - a2x1 + K1rx.3 = x4x.4 = - b2x3 + K2ry = x1x350.The equivalent forward transfer function is G(s) =K1K2s(s+a1) . The equivalent feedback transfer functionisH(s) = K3 +K4ss+a2. Hence, the closed-loop transfer function isT(s) =G(s)1 + G(s)H(s) =K1K2(s+a2)s3 + (a1+a2)s2 + (a1a2+K1K2K3+K1K2K4)s + K1K2K3a251.a. The equivalent forward transfer function isG e s K5s s 2+1 5s s 2++1s s 3+= = 5 Ks s 3+ s 2 2 s 5+ +T sG e1 G e+= = 5 Ks 4 5 s 3 11 s 2 15 s 5 K+ + + +b. Draw the signal-flow diagram:3x4xyux 12xK 51s-21s-111s1s-3-11 1Writing the state and output equations from the signal-flow diagram:
  • 251. 5-64 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.x1.= x2x2.= −3x2 + x3x3.= x4x4.= −5Kx1 − 5x3 − 2x4 +5Kuy = x1In vector-matrix form:x.=0 1 0 00 −3 1 00 0 0 1−5K 0 −5 −2⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥x +0005K⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥uy = 1 0 0 0[ ]xc.Program:for K=1:1:5numt=5*K;dent=[1 5 11 15 5*K];T=tf(numt,dent);hold on;subplot(2,3,K);step(T,0:0.01:20)title([K=,int2str(K)])endComputer response:
  • 252. Solutions to Problems 5-65Copyright © 2011 by John Wiley & Sons, Inc.52.a. Draw the signal-flow diagram:3x4xyux12x11s1666.67 0.06-72015x10 6-2x10 7-4x10 6-82-11s1s1s 1Write state and output equations from the signal-flow diagram:x1..= x2x2.= x3x3.= −2*107x1 − 4 *106x2 − 82x3 +15*106x4x4.= −100x1 − 720x4 +100uy = x1
  • 253. 5-66 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.In vector-matrix form:x.=0 1 0 00 0 1 0−2*107−4 *106−82 15*106−100 0 0 −720⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥x +000100⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥uy = 1 0 0 0[ ]xb.Program:numg=1666.67*0.06*15e6;deng=conv([1 720],[1 82 4e6 2e7]);G(s)G=tf(numg,deng)T(s)T=feedback(G,1)step(T)Computer response:ans =G(s)Transfer function:1.5e009----------------------------------------------------s^4 + 802 s^3 + 4.059e006 s^2 + 2.9e009 s + 1.44e010ans =T(s)Transfer function:1.5e009----------------------------------------------------s^4 + 802 s^3 + 4.059e006 s^2 + 2.9e009 s + 1.59e010
  • 254. Solutions to Problems 5-67Copyright © 2011 by John Wiley & Sons, Inc.53.a. Phase-variable from: G(s) =-272(s2+1.9s+84)s3+17.1s2+34.58s-123.48Drawing the signal-flow diagram:84y3xu x 12x1s-272123.48-34.58-17.111.91s1sWriting the state and output equations:x1..= x2x2.= x3x3.= 123.48x1 − 34.58x2 −17.1x3 − 272uy = 84x1 +1.9x2 + x3In vector-matrix form:x.=0 1 00 0 1123.48 −34.58 −17.1⎡⎣⎢⎢⎤⎦⎥⎥x +00−272⎡⎣⎢⎢⎤⎦⎥⎥uy = 84 1.9 1[ ]xb. Controller canonical form: G(s) =-272(s2+1.9s+84)s3+17.1s2+34.58s-123.48Drawing the signal-flow diagram:
  • 255. 5-68 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.84y3xu x 1 2x1s-272123.48-34.58-17.111.91s1sWriting the state and output equations:x1.= −17.1x1 −34.58x2 +123.48x3 − 272ux2.= x1..x3.= x2y = x1 +1.9x2 + 84x3In vector-matrix form:x.=−17.1 −34.58 123.481 0 00 1 0⎡⎣⎢⎢⎤⎦⎥⎥x +−27200⎡⎣⎢⎢⎤⎦⎥⎥uy = 1 1.9 84[ ]xc. Observer canonical form: Divide by highest power of s and obtainG(s) =-272s -516.8s2 -22848s31 +17.1s +34.58s2 -123.48s3Cross multiplying,[-272s -516.8s2 -22848s3 ]R(s) = [ 1 +17.1s +34.58s2 -123.48s3 ]C(s)Rearranging,C(s) =1s [ -272R(s) - 17.1C(s)] +1s2 [ -516.8R(s) - 34.58C(s)] +1s3 [ -22848R(s) + 123.48C(s)]Drawing the signal-flow diagram, where r = u and y = c:
  • 256. Solutions to Problems 5-69Copyright © 2011 by John Wiley & Sons, Inc.yu1s1s1s13x x 12x1122848-516.8123.48-34.58-17.1-272d. Draw signal-flow ignoring the polynomial in the numerator:u1s1s1s13x x 12x1-272-14 1.8 -4.9Write the state equations:x1.= −4.9x1 + x2x2.=1.8x2 + x3..x3.= −14x3 −272uThe output equation isy = x1..+1.9 x1.+84x1 (1)But,x1.= −4.9x1 + x2 (2)andx1..= −4.9 x1.+ x2.= −4.9(−4.9x1 + x2 ) +1.8x2 + x3 (3)
  • 257. 5-70 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.Substituting Eqs. (2) and (3) into (1) yields,y = 98.7x1 −1.2x2 + x3In vector-matrix form:x.=−4.9 1 00 1.8 10 0 −14⎡⎣⎢⎢⎤⎦⎥⎥x +00−272⎡⎣⎢⎢⎤⎦⎥⎥uy = 98.7 −1.2 1[ ]xe. Expand as partial fractions:G s 479.38 1s 14+− 232.94 1s 1.8−− 440.32 1s 4.9++=Draw signal-flow diagram:-141.811113x1syux 12x1s1s-479.38-232.94440.32-4.9Write state and output equations:x1.= −14x1 + −479.38ux2.=1.8x2 − 232.94u..x3.= −4.9x3 + 440.32uy = x1 + x2 + x3In vector-matrix form:x.=−14 0 00 1.8 00 0 −4.9⎡⎣⎢⎢⎤⎦⎥⎥x +−479.38−232.94440.32⎡⎣⎢⎢⎤⎦⎥⎥uy = 1 1 1[ ]x
  • 258. Solutions to Problems 5-71Copyright © 2011 by John Wiley & Sons, Inc.54.Push Pitch Gain to the right past the pickoff point.Collapse the summing junctions and add the feedback transfer functions.Apply the feedback formula and obtain,T(s) =G(s)1 + G(s)H(s)=0.25(s + 0.435)s4+ 3.4586s3+ 3.4569s2+ 0.9693s + 0.1503255.Program:numg1=-0.125*[1 0.435]deng1=conv([1 1.23],[1 0.226 0.0169])G1G1=tf(numg1,deng1)G2G2=tf(2,[1 2])G3=-1H1H1=tf([-1 0],1)Inner LoopGe=feedback(G1*G2,H1)Closed-LoopT=feedback(G3*Ge,1)Computer response:numg1 =-0.1250 -0.0544
  • 259. 5-72 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.deng1 =1.0000 1.4560 0.2949 0.0208ans =G1Transfer function:-0.125 s - 0.05438------------------------------------s^3 + 1.456 s^2 + 0.2949 s + 0.02079ans =G2Transfer function:2-----s + 2G3 =-1ans =H1Transfer function:-sans =Inner LoopTransfer function:-0.25 s - 0.1088------------------------------------------------s^4 + 3.456 s^3 + 3.457 s^2 + 0.7193 s + 0.04157ans =Closed-LoopTransfer function:0.25 s + 0.1088-----------------------------------------------s^4 + 3.456 s^3 + 3.457 s^2 + 0.9693 s + 0.1503
  • 260. Solutions to Problems 5-73Copyright © 2011 by John Wiley & Sons, Inc.56.
  • 261. 5-74 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.57.a. Since VL(s) = Vg(s) – VR(s), the summing junction has Vg(s) as the positive input and VR(s) as the negativeinput, and VL(s) as the error. Since I(s) = VL(s) (1/(Ls)), G(s) = 1/(Ls). Also, since VR(s) = I(s)R, the feedback isH(s) = R. Summarizing, the circuit can be modeled as a negative feedback system, where G(s) = 1/(Ls), H(s) =R, input = Vg(s), output = I(s), and error = VL(s), where the negative input to the summing junction is VR(s).b. T(s) =I(s)Vg(s)=G(s)1 + G(s)H(s)=1Ls1 +1LsR=1Ls + R. Hence, I(s) = Vg (s)1Ls + R.c. Using circuit analysis, I(s) =Vg (s)Ls + R.58.a.21sTαβ= ,sLα−=1 ,sLβ−=2 , No non-touching loops.ssβα++=Δ 1 , 11 =ΔLinearDeadzoneLinearBacklash
  • 262. Solutions to Problems 5-75Copyright © 2011 by John Wiley & Sons, Inc.[ ])(1)(211βααββααβ++=++=ΔΔ=sssssTsHHrmb.sTα=1 ,sLα−=1 ,sLβ−=2 , No non-touching loops.ssβα++=Δ 1 , 11 =Δ)(1)( 11βααβαα++=++=ΔΔ=ssssTsHHrfc.α=1T ,sLα−=1 ,sLβ−=2 , No non-touching loops.ssβα++=Δ 1 , 11 =Δ)(1)( 11βααβαα++=++=ΔΔ=ssssTsHQrid.sTαβ=1 ,sLα−=1 ,sLβ−=2 , No non-touching loops.ssβα++=Δ 1 , 11 =Δ)(1)( 11βααββααβ++=++=ΔΔ=ssssTsHQroe.LetsKsHr =)( . From part d) KsHssLimssQLimq rsosoβααββααβ+=++==∞→→)()()()(00=constantFrom (a)[ ] [ ]trmmeKKtKsKsKsKssKsHsssHth)(32132212)()()()()}({)(βαβαβααββααβ+−++=⎭⎬⎫⎩⎨⎧++++=⎭⎬⎫⎩⎨⎧++=⎭⎬⎫⎩⎨⎧++==1-1-1-1-LLLL
  • 263. 5-76 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.So 21)( KtKthLim mt+=∞→. It is clear that )(thm increases at constant speed.59.a.We start by finding1UUand2UUusing Mason’s Rule1UUpcGGT =1 ; Loops: vl GGALL 221 == , no non-touching loops. vl GGA221−=Δ , 11 =Δ FollowsvlpCGGAGGUU21 21−=2UUpcGGT −=1 ; Loops: vl GGALL 221 == , no non-touching loops. vl GGA221−=Δ , 11 =Δ FollowsvlpCGGAGGUU22 21−−=By superposition )(21212UUGGAGGUvlpC−−=b.Now we findextFUTwo forward paths: vl GGATT 221 == Loops: vl GGALL 221 == , no non-touching loops.vl GGA221−=Δ . 121 =Δ=Δvlvlext GGAGGAFU22212−=c. From (a) to get 0=u it is required that 21 uu = .60.a.The first equation follows from the schematic. The second equation is obtained by applying the voltagedivider rule at the op-amp’s inverting terminal, noting that since the op-amp considered is ideal, there is nocurrent demand there.b. AT =1 ;fiiRRRAL+−= ;fiiRRRA++=Δ 1 ; 11 =Δ
  • 264. Solutions to Problems 5-77Copyright © 2011 by John Wiley & Sons, Inc.fiiioRRRAATvv++=ΔΔ=111c.iffiifiiAioRRRRRRRRAALimvv+=+=++=∞→11161.a.Adding currents at the op-amp’s inverting terminal, under ideal condition we getfoiiRvvRvv −=− 11which after some algebraic manipulations gives oifiiiffvRRRvRRRv+++=1Also from the circuits diagram 1Avvo −=b.These equations can be represented by the following block diagramWe have thatiffRRRAT+−=1 ;fiiRRRAL+−= ;fiiRRRA++=Δ 1 ; 11 =ΔfiififioRRRARRRATvv+++−=ΔΔ=111c.iffiififfiififAioRRRRRRRRRRRARRRALimvv=++−=+++−=∞→162.a.The three equations follow by direct observation from the small signal circuit.b.The block diagram is given by- AifiRRR+iffRRR+++vi Vo)||( osm rRgsiiRRR+VoVi+-
  • 265. 5-78 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.c. From the block diagram we getsiiosmosmioRRRrRgrRgvv++=)||(1)||(63.a. Using Mason’s rule2022020211ωωω +=+=sMsKTUSt; Loops rmssLεεωω+++= 202201 andεε+−=smL r2 , no non-touching loops. 11 =Δ))((11 2022202202022020220113ωεεωωεεωωεεωω++++=++−+++=ΔΔ=sssmsmsssmsTRXrrra. From part (a)))((1 20222022031ωεεεεωωεε+++++=+=sssmsssRXRXr64.a.>> A=[-100.2 -20.7 -30.7 200.3; 40 -20.22 49.95 526.1;...0 10.22 -59.95 -526.1; 0 0 0 0];>> B=[208; -208; -108.8; -1];>> C = [0 1570 1570 59400];>> D = -6240;>> [n,d]=ss2tf(A,B,C,D)
  • 266. Solutions to Problems 5-79Copyright © 2011 by John Wiley & Sons, Inc.n =1.0e+009 *Columns 1 through 3-0.00000624000000 -0.00168228480000 -0.14206098728000Columns 4 through 5-3.91955218234127 -9.08349454230472d =1.0e+005 *Columns 1 through 30.00001000000000 0.00180370000000 0.09562734000000Columns 4 through 51.32499100000000 0
  • 267. 5-80 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.>> roots(n)ans =1.0e+002 *-1.34317654991673-0.78476212102923-0.54257777928519-0.02545278053809>> roots(d)ans =0-92.38329312886714-66.38046756013043-21.60623931100260Note that 14.68555)0( =UYs , follows that)4.92)(4.66)(6.21()3.134)(5.78)(3.54)(5.2(17.6348)(+++++++−=sssssssssUYb.>> [r,p,k]=residue(n,d)
  • 268. Solutions to Problems 5-81Copyright © 2011 by John Wiley & Sons, Inc.r =1.0e+005 *-0.73309459854184-0.51344619392820-3.63566779304453-0.68555141448543p =-92.38329312886714-66.38046756013043-21.606239311002600k =-6240or4.9214.685554.668.3635666.216.5134446.733096240)(+−+−+−−−=sssssUY
  • 269. 5-82 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.c.d. The corresponding state space representation is:)(14.685558.3635666.513445.733094.900004.6600006.210000043214321tuxxxxxxxx⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−+⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡&&&&[ ] )(624011114321tuxxxxy −⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=65.a.>> A = [0 1 0; 0 -68.3 -7.2; 0 3.2 -0.7]
  • 270. Solutions to Problems 5-83Copyright © 2011 by John Wiley & Sons, Inc.A =0 1.0000 00 -68.3000 -7.20000 3.2000 -0.7000>> [V,D]=eig(A)V =1.0000 0.0147 -0.10160 -0.9988 0.10590 0.0475 -0.9892D =0 0 00 -67.9574 00 0 -1.0426Matrix V is the sought similarity transformation.
  • 271. 5-84 Chapter 5: Reduction of Multiple SubsystemsCopyright © 2011 by John Wiley & Sons, Inc.b.>> Ad = inv(V)*A*VAd =0 -0.0000 -0.00000 -67.9574 0.00000 -0.0000 -1.0426>> B = [0;425.4;0]B =0425.40000>> Bd = inv(V)*BBd =4.2030-428.1077-20.5661The diagonalized system is:
  • 272. Solutions to Problems 5-85Copyright © 2011 by John Wiley & Sons, Inc.mezzzzzz⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡57.2011.4282.40426.100096.670000321321&&&66.a.b.There